Giải phương trình: \(\sin 2 x=-\displaystyle\frac{1}{2}\).
\(\begin{aligned}&\sin 2 x=-\displaystyle\frac{1}{2}\\ \Leftrightarrow\ &\sin 2 x=\sin\left( -\displaystyle\frac{\pi}{6}\right) \\ \Leftrightarrow\ &\left[\begin{aligned}&2x=-\displaystyle\frac{\pi}{6}+k2\pi\\ &2x=\pi+\displaystyle\frac{\pi}{6}+k2\pi\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=-\displaystyle\frac{\pi}{12}+k\pi\\ &x=\displaystyle\frac{7\pi}{12}+k\pi\end{aligned}\right. (k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace -\displaystyle\frac{\pi}{12}+k\pi;\displaystyle\frac{7\pi}{12}+k\pi\ \Big | \, k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\sin 2 x=\displaystyle\frac{\sqrt{2}}{2}\).
\(\begin{aligned}&\sin 3x=\displaystyle\frac{\sqrt{2}}{2}\\ \Leftrightarrow\ &\sin 3x=\sin\left( \displaystyle\frac{\pi}{4}\right) \\ \Leftrightarrow\ &\left[\begin{aligned}&3x=\displaystyle\frac{\pi}{4}+k2\pi\\ &3x=\pi-\displaystyle\frac{\pi}{4}+k2\pi\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=\displaystyle\frac{\pi}{12}+k\displaystyle\frac{2\pi}{3}\\ &x=\displaystyle\frac{\pi}{4}+k\displaystyle\frac{2\pi}{3}\end{aligned}\right. (k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{\pi}{12}+k\displaystyle\frac{2\pi}{3};\displaystyle\frac{\pi}{4}+k\displaystyle\frac{2\pi}{3}\ \Big | \, k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(2\sin\left( 2x-\displaystyle\frac{\pi}{6}\right)=1\).
\(\begin{aligned}&2\sin\left( 2x-\displaystyle\frac{\pi}{6}\right)=1\\ \Leftrightarrow\ &\sin\left( 2x-\displaystyle\frac{\pi}{6}\right)=\displaystyle\frac{1}{2}\\ \Leftrightarrow\ &\sin\left( 2x-\displaystyle\frac{\pi}{6}\right)=\sin\displaystyle\frac{\pi}{6}\\ &\\ \Leftrightarrow\ &\left[\begin{aligned}&2x-\displaystyle\frac{\pi}{6}=\displaystyle\frac{\pi}{6}+k2\pi\\ &2x-\displaystyle\frac{\pi}{6}=\pi-\displaystyle\frac{\pi}{6}+k2\pi\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=\displaystyle\frac{\pi}{6}+k\pi\\ &x=\displaystyle\frac{\pi}{2}+k\pi\end{aligned}\right. (k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{\pi}{6}+k\pi;\displaystyle\frac{\pi}{2}+k\pi\ \Big | \, k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(2 \sin \left( 3x-\displaystyle\frac{\pi}{4}\right) =\sqrt{3}\).
\(\begin{aligned}&2 \sin \left( 3x-\displaystyle\frac{\pi}{4}\right) =\sqrt{3}\\ \Leftrightarrow\ &\sin \left( 3x-\displaystyle\frac{\pi}{4}\right) =\displaystyle\frac{\sqrt{3}}{2}\\ \Leftrightarrow\ &\sin \left( 3x-\displaystyle\frac{\pi}{4}\right)=\sin\displaystyle\frac{\pi}{3}\\ &\\ \Leftrightarrow\ &\left[\begin{aligned}&3x-\displaystyle\frac{\pi}{4}=\displaystyle\frac{\pi}{3}+k2\pi\\ &3x-\displaystyle\frac{\pi}{4}=\pi-\displaystyle\frac{\pi}{3}+k2\pi\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=\displaystyle\frac{7\pi}{36}+k\displaystyle\frac{2\pi}{3}\\ &x=\displaystyle\frac{11\pi}{36}+k\displaystyle\frac{2\pi}{3}\end{aligned}\right. (k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{7\pi}{36}+k\displaystyle\frac{2\pi}{3};\displaystyle\frac{11\pi}{36}+k\displaystyle\frac{2\pi}{3}\ \Big | \, k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(2 \sin \left(x-20^{\circ}\right)-\sqrt{3}=0\).
\(\begin{aligned}&2 \sin \left(x-20^{\circ}\right)-\sqrt{3}=0&\\ \Leftrightarrow\ &\sin \left(x-20^{\circ}\right)=\displaystyle\frac{\sqrt{3}}{2}\\ \Leftrightarrow\ &\sin \left(x-20^{\circ}\right)=\sin 60^\circ\\ \Leftrightarrow\ &\left[\begin{aligned}&x-20^\circ=60^\circ+k360^\circ\\ &x-20^\circ=180^\circ-60^\circ+k360^\circ\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=80^\circ+k360^\circ\\ &x=140^\circ+k360^\circ\end{aligned}\right.\ (k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace 80^\circ+k360^\circ;140^\circ+k360^\circ\ \Big | \, k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\sin \left(4 x+45^{\circ}\right)=\sin 2x\).
\(\begin{aligned}&\sin \left(4 x+45^{\circ}\right)=\sin 2 x&\\ \Leftrightarrow\ &\left[\begin{aligned}&4 x+45^{\circ}=2x+k360^\circ\\ &4 x+45^{\circ}=180^\circ-2x+k360^\circ\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=-22{,}5^\circ+k180^\circ\\ &x=22{,}5^\circ+k60^\circ\end{aligned}\right.(k\in\mathbb{Z}).\end{aligned}\)
\(\begin{aligned}&\sin \left(4 x+45^{\circ}\right)=\sin 2 x&\\ \Leftrightarrow\ &\left[\begin{aligned}&4 x+45^{\circ}=2x+k360^\circ\\ &4 x+45^{\circ}=180^\circ-2x+k360^\circ\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=-22{,}5^\circ+k180^\circ\\ &x=22{,}5^\circ+k60^\circ\end{aligned}\right.\ (k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace-22{,}5^\circ+k180^\circ;22{,}5^\circ+k60^\circ\ \Big | \, k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\sin 3 x=-1\).
\(\begin{aligned}&\sin 3 x=-1\\ \Leftrightarrow\ & 3x=-\displaystyle\frac{\pi}{2}+k2\pi\\ \Leftrightarrow\ & x=-\displaystyle\frac{\pi}{6}+k\displaystyle\frac{2\pi}{3}\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{\pi}{6}+k\displaystyle\frac{2\pi}{3}\ \Big | \, k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\sin 3x=0\).
\(\begin{aligned}&\sin 3x=0\\ \Leftrightarrow\ & 3x=k\pi\\ \Leftrightarrow\ & x=k\displaystyle\frac{\pi}{3}\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace k\displaystyle\frac{\pi}{3}\ \Big | \, k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(2 \cos 2 x=-\sqrt{2}\).
\(\begin{aligned}&2\cos 2 x=-\sqrt{2}\\ \Leftrightarrow\ & \cos2x=-\displaystyle\frac{\sqrt{2}}{2}\\ \Leftrightarrow\ &\cos2x=\cos\displaystyle\frac{3\pi}{4}\\ \Leftrightarrow\ &2x=\pm\displaystyle\frac{3\pi}{4}+k2\pi\\ \Leftrightarrow\ & x=\pm\displaystyle\frac{3\pi}{8}+k\pi\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \pm\displaystyle\frac{3\pi}{8}+k\pi\ \Big | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(2 \cos \left( 2x-\displaystyle\frac{\pi}{3}\right) +\sqrt{3}=0\).
\(\begin{aligned}&2 \cos \left( 2x-\displaystyle\frac{\pi}{3}\right) +\sqrt{3}=0&\\ \Leftrightarrow\ &\cos \left( 2x-\displaystyle\frac{\pi}{3}\right)=-\displaystyle\frac{\sqrt{3}}{2}\\ \Leftrightarrow\ &\cos \left( 2x-\displaystyle\frac{\pi}{3}\right)=\cos\displaystyle\frac{5\pi}{6}\\ \Leftrightarrow\ &\left[\begin{aligned}&2x-\displaystyle\frac{\pi}{3}=\displaystyle\frac{5\pi}{6}+k2\pi\\ &2x-\displaystyle\frac{\pi}{3}=-\displaystyle\frac{5\pi}{6}+k2\pi\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=\displaystyle\frac{7\pi}{12}+k\pi\\ &x=-\displaystyle\frac{\pi}{4}+k\pi\end{aligned}\right.\ (k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{7\pi}{12}+k\pi;-\displaystyle\frac{\pi}{4}+k\pi\ \Big | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(2 \cos \left(3 x+15^{\circ}\right)-\sqrt{3}=0\).
\(\begin{aligned}&2 \cos \left(3 x+15^{\circ}\right)-\sqrt{3}=0\\ \Leftrightarrow\ &\cos \left(3 x+15^{\circ}\right)=\displaystyle\frac{\sqrt{3}}{2}\\ \Leftrightarrow\ &\cos \left(3 x+15^{\circ}\right)=\cos30^\circ\\ \Leftrightarrow\ &\left[\begin{aligned}&3x+15^\circ=30^\circ+k360^\circ\\ &3x+15^\circ=-30^\circ+k360^\circ\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=5^\circ+k120^\circ\\ &x=-15^\circ+k120^\circ\end{aligned}\right.\ (k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace 5^\circ+k120^\circ;-15^\circ+k120^\circ\ \Big | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(2 \cos \left(x-55^{\circ}\right)=\sqrt{2}\).
\(\begin{aligned}&2 \cos \left(x-55^{\circ}\right)=\sqrt{2}&\\ \Leftrightarrow\ &\cos \left(x-55^{\circ}\right)=\displaystyle\frac{\sqrt{2}}{2}\\ \Leftrightarrow\ &\cos \left(x-55^{\circ}\right)=\cos45^\circ\\ \Leftrightarrow\ &\left[\begin{aligned}&x-55^{\circ}=45^\circ+k360^\circ\\ &x-55^{\circ}=-45^\circ+k360^\circ\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=100^\circ+k360^\circ\\ &x=10^\circ+k360^\circ\end{aligned}\right.\ (k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace 100^\circ+k360^\circ;10^\circ+k360^\circ\ \Big | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\tan \left(x+\displaystyle\frac{\pi}{12}\right)=\displaystyle\frac{\sqrt{3}}{3}\).
\(\begin{aligned}&\tan \left(x+\displaystyle\frac{\pi}{12}\right)=\displaystyle\frac{\sqrt{3}}{3}\\ \Leftrightarrow\ &\tan \left(x+\displaystyle\frac{\pi}{12}\right)=\tan \displaystyle\frac{\pi}{6}\\ \Leftrightarrow\ & x+\displaystyle\frac{\pi}{12}=\displaystyle\frac{\pi}{6}+k\pi\\ \Leftrightarrow\ & x=\displaystyle\frac{\pi}{12}+k\pi\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{\pi}{12}+k\pi\ \Big | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\tan \left(2 x-\displaystyle\frac{\pi}{6}\right)=\sqrt{3}\).
\(\begin{aligned}&\tan \left(2 x-\displaystyle\frac{\pi}{6}\right)=\sqrt{3}\\ \Leftrightarrow\ &\tan \left(2 x-\displaystyle\frac{\pi}{6}\right)=\tan \displaystyle\frac{\pi}{3}\\ \Leftrightarrow\ & 2 x-\displaystyle\frac{\pi}{6}=\displaystyle\frac{\pi}{3}+k\pi\\ \Leftrightarrow\ & 2x=\displaystyle\frac{\pi}{2}+k\pi\\ \Leftrightarrow\ &x=\displaystyle\frac{\pi}{4}+k\displaystyle\frac{\pi}{2}\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{\pi}{4}+k\displaystyle\frac{\pi}{2}\ \Big | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(3 \tan \left( 2x+\displaystyle\frac{\pi}{6}\right) =-\sqrt{3}\).
\(\begin{aligned}&3 \tan \left( 2x+\displaystyle\frac{\pi}{6}\right) =-\sqrt{3}\\ \Leftrightarrow\ &\tan \left(2x+\displaystyle\frac{\pi}{6}\right)=-\displaystyle\frac{\sqrt{3}}{3}\\ \Leftrightarrow\ &\tan \left(2x+\displaystyle\frac{\pi}{6}\right)=\tan \left( -\displaystyle\frac{\pi}{6}\right) \\ \Leftrightarrow\ & 2x+\displaystyle\frac{\pi}{6}=-\displaystyle\frac{\pi}{6}+k\pi\\ \Leftrightarrow\ & 2x=-\displaystyle\frac{\pi}{3}+k\pi\\ \Leftrightarrow\ &x=-\displaystyle\frac{\pi}{6}+k\displaystyle\frac{\pi}{2}\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace -\displaystyle\frac{\pi}{6}+k\displaystyle\frac{\pi}{2}\ \Big | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\tan \left(x-\displaystyle\frac{\pi}{4} \right) +\sqrt{3}=0\).
\(\begin{aligned}&\tan \left(x-\displaystyle\frac{\pi}{4} \right) +\sqrt{3}=0\\ \Leftrightarrow\ &\tan \left(x-\displaystyle\frac{\pi}{4}\right)=-\sqrt{3}\\ \Leftrightarrow\ &\tan \left(x-\displaystyle\frac{\pi}{4}\right)=\tan \left( -\displaystyle\frac{\pi}{3}\right) \\ \Leftrightarrow\ & x-\displaystyle\frac{\pi}{4}=-\displaystyle\frac{\pi}{3}+k\pi\\ \Leftrightarrow\ & x=-\displaystyle\frac{\pi}{12}+k\pi\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace -\displaystyle\frac{\pi}{12}+k\pi\ \Big | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\cot \left(3 x-\displaystyle\frac{\pi}{4}\right)=\sqrt{3}\).
\(\begin{aligned}&\cot \left(3 x-\displaystyle\frac{\pi}{4}\right)=\sqrt{3}\\ \Leftrightarrow\ &\cot \left(3x-\displaystyle\frac{\pi}{4}\right)=\sqrt{3}\\ \Leftrightarrow\ &\cot \left(3x-\displaystyle\frac{\pi}{4}\right)=\cot \displaystyle\frac{\pi}{6} \\ \Leftrightarrow\ & 3x-\displaystyle\frac{\pi}{4}=\displaystyle\frac{\pi}{6}+k\pi\\ \Leftrightarrow\ &3x=\displaystyle\frac{5\pi}{12}+k\pi\\ \Leftrightarrow\ & x=\displaystyle\frac{5\pi}{36}+k\displaystyle\frac{\pi}{3}\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{5\pi}{36}+k\displaystyle\frac{\pi}{3}\ \Big | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\cot \left(\displaystyle\frac{x}{2}-\displaystyle\frac{\pi}{5}\right)=\displaystyle\frac{\sqrt{3}}{3}\).
\(\begin{aligned}&\cot \left(\displaystyle\frac{x}{2}-\displaystyle\frac{\pi}{5}\right)=\displaystyle\frac{\sqrt{3}}{3}\\ \Leftrightarrow\ &\cot \left(3x-\displaystyle\frac{\pi}{4}\right)=\displaystyle\frac{\sqrt{3}}{3}\\ \Leftrightarrow\ &\cot \left(\displaystyle\frac{x}{2}-\displaystyle\frac{\pi}{5}\right)=\cot \displaystyle\frac{\pi}{3} \\ \Leftrightarrow\ & \displaystyle\frac{x}{2}-\displaystyle\frac{\pi}{5}=\displaystyle\frac{\pi}{3}+k\pi\\ \Leftrightarrow\ &\displaystyle\frac{x}{2}=\displaystyle\frac{8\pi}{15}+k\pi\\ \Leftrightarrow\ & x=\displaystyle\frac{16\pi}{15}+k2\pi\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{16\pi}{15}+k2\pi\ \Big | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\cot \left(x+\displaystyle\frac{\pi}{3} \right) =-1\).
\(\begin{aligned}&\cot \left(x+\displaystyle\frac{\pi}{3} \right) =-1\\ \Leftrightarrow\ &x+\displaystyle\frac{\pi}{3}=-\displaystyle\frac{\pi}{4}+k\pi\\ \Leftrightarrow\ &x=-\displaystyle\frac{7\pi}{12}+k\pi\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace -\displaystyle\frac{7\pi}{12}+k\pi\ \Big | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\tan \left(x+\displaystyle\frac{\pi}{3} \right)=1\).
\(\begin{aligned}&\tan \left(x+\displaystyle\frac{\pi}{3} \right)=1\\ \Leftrightarrow\ &x+\displaystyle\frac{\pi}{3}=\displaystyle\frac{\pi}{4}+k\pi\\ \Leftrightarrow\ &x=-\displaystyle\frac{\pi}{12}+k\pi\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace -\displaystyle\frac{\pi}{12}+k\pi\ \Big | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\left( \sqrt{3} \tan x+3\right) (2 \cos 2 x+1)=0\).
\(\begin{aligned}&\left(\sqrt{3} \tan x+3\right) (2 \cos 2 x+1)=0\\ \Leftrightarrow\ &\left[\begin{aligned}&\sqrt{3} \tan x+3=0\\&2 \cos 2x+1=0\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&\tan x=-\sqrt{3}\\&\cos 2x=-\displaystyle\frac{1}{2}\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&\tan x=\tan \left( -\displaystyle\frac{\pi}{3}\right) \\&\cos 2x=\cos \displaystyle\frac{2\pi}{3}\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=-\displaystyle\frac{\pi}{3}+k\pi\\&2x=\pm\displaystyle\frac{2\pi}{3}+k2\pi\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=-\displaystyle\frac{\pi}{3}+k\pi\\&x=\pm\displaystyle\frac{\pi}{3}+k\pi\end{aligned}\right.\\ \Leftrightarrow\ &x=\pm\displaystyle\frac{\pi}{3}+k\pi\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \pm\displaystyle\frac{\pi}{3}+k\pi\ | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\left( \sqrt{2} \sin 2 x+2\right) \left( 2 \cos x+\sqrt{2}\right) =0\).
\(\begin{aligned}&\left( \sqrt{2} \sin 2 x+2\right) \left( 2 \cos x+\sqrt{2}\right) =0\\ \Leftrightarrow\ &\left[\begin{aligned}&\sqrt{2} \sin 2 x+2=0\\&2 \cos x+\sqrt{2}=0\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&\sin 2x=-\sqrt{2}\text{ (phương trình vô nghiệm)}\\&\cos x=-\displaystyle\frac{\sqrt{2}}{2}\end{aligned}\right.\\ \Leftrightarrow\ &\cos x=\cos \displaystyle\frac{3\pi}{4}\\ \Leftrightarrow\ &x=\pm\displaystyle\frac{3\pi}{4}+k2\pi\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \pm\displaystyle\frac{3\pi}{4}+k\pi\ | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \((\sin x+1)\left( 2 \cos 2 x-\sqrt{2}\right) =0\).
\(\begin{aligned}&(\sin x+1)\left( 2 \cos 2 x-\sqrt{2}\right) =0\\ \Leftrightarrow\ &\left[\begin{aligned}&\sin x+1=0\\& 2 \cos 2 x-\sqrt{2}=0\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&\sin x=-1\\&\cos 2x=\displaystyle\frac{\sqrt{2}}{2}\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=-\displaystyle\frac{\pi}{2}+k2\pi\\&\cos 2x=\cos\displaystyle\frac{\pi}{4}\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=-\displaystyle\frac{\pi}{2}+k2\pi\\&x=\pm\displaystyle\frac{\pi}{8}+k\pi\end{aligned}\right.(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace -\displaystyle\frac{\pi}{2}+k2\pi;\pm\displaystyle\frac{\pi}{8}+k\pi\ | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \((2 \sin x-1)\left( \sqrt{3} \cos x-5\right) =0\).
\(\begin{aligned}&(2 \sin x-1)\left( \sqrt{3} \cos x-5\right) =0\\ \Leftrightarrow\ &\left[\begin{aligned}&2 \sin x-1=0\\& \sqrt{3} \cos x-5=0\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&\sin x=\displaystyle\frac{1}{2}\\&\cos x=\displaystyle\frac{5\sqrt{3}}{3}\text{ (phương trình vô nghiệm)}\end{aligned}\right.\\ \Leftrightarrow\ &\sin x=\sin\displaystyle\frac{\pi}{6}\\ \Leftrightarrow\ &\left[\begin{aligned}&x=\displaystyle\frac{\pi}{6}+k2\pi\\&x=\displaystyle\frac{5\pi}{6}+k2\pi\end{aligned}\right.(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{\pi}{6}+k2\pi;\displaystyle\frac{5\pi}{6}+k2\pi\ | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\sin 3 x \cos 3 x-\displaystyle\frac{\sqrt{2}}{4}=0\).
\(\begin{aligned}&\sin 3 x \cos 3 x-\displaystyle\frac{\sqrt{2}}{4}=0\\ \Leftrightarrow\ &\sin6x=\displaystyle\frac{\sqrt{2}}{2}\\ \Leftrightarrow\ &\sin6x=\sin\displaystyle\frac{\pi}{4}\\ \Leftrightarrow\ &\left[\begin{aligned}&6x=\displaystyle\frac{\pi}{4}+k2\pi\\&6x=\pi-\displaystyle\frac{\pi}{4}+k2\pi\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=\displaystyle\frac{\pi}{24}+k\displaystyle\frac{\pi}{3}\\&x=\displaystyle\frac{\pi}{8}+k\displaystyle\frac{\pi}{3}\end{aligned}\right.(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{\pi}{24}+k\displaystyle\frac{\pi}{3};\displaystyle\frac{\pi}{8}+k\displaystyle\frac{\pi}{3}\ | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\sin x \cos x \cos 2 x=\displaystyle\frac{1}{4}\).
\(\begin{aligned}&\sin x \cos x \cos 2 x=\displaystyle\frac{1}{4}\\ \Leftrightarrow\ &\displaystyle\frac{1}{2}\sin 2x \cos 2 x=\displaystyle\frac{1}{4}\\ \Leftrightarrow\ &\displaystyle\frac{1}{4}\sin4x=\displaystyle\frac{1}{4}\\ \Leftrightarrow\ &\sin4x=1\\ \Leftrightarrow\ &4x=\displaystyle\frac{\pi}{2}+k2\pi\\ \Leftrightarrow\ &x=\displaystyle\frac{\pi}{8}+k\displaystyle\frac{\pi}{2}\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{\pi}{8}+k\displaystyle\frac{\pi}{2}\ | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\sin 2 x \cdot \cos 2 x \cdot \cos 4 x+\displaystyle\frac{1}{8}=0\).
\(\begin{aligned}&\sin 2 x \cdot \cos 2 x \cdot \cos 4 x+\displaystyle\frac{1}{8}=0\\ \Leftrightarrow\ &\displaystyle\frac{1}{2}\sin 4x \cdot \cos 4 x=-\displaystyle\frac{1}{8}\\ \Leftrightarrow\ &\displaystyle\frac{1}{4}\sin8x=-\displaystyle\frac{1}{8}\\ \Leftrightarrow\ &\sin8x=-\displaystyle\frac{1}{2}\\ \Leftrightarrow\ &\sin8x=\sin \left( -\displaystyle\frac{\pi}{6}\right)\\ \Leftrightarrow\ &\left[\begin{aligned}&8x=-\displaystyle\frac{\pi}{6}+k2\pi\\&8x=\pi+\displaystyle\frac{\pi}{6}+k2\pi\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=-\displaystyle\frac{\pi}{48}+k\displaystyle\frac{\pi}{4}\\&x=\displaystyle\frac{7\pi}{48}+k\displaystyle\frac{\pi}{4}\end{aligned}\right.(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace -\displaystyle\frac{\pi}{48}+k\displaystyle\frac{\pi}{4};\displaystyle\frac{7\pi}{48}+k\displaystyle\frac{\pi}{4}\ | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\sin x \cos x \cos 2 x \cos 4 x \cos 8 x=\displaystyle\frac{1}{16}\).
\(\begin{aligned}&\sin x \cos x \cos 2 x \cos 4 x \cos 8 x=\displaystyle\frac{1}{16}\\ \Leftrightarrow\ &\displaystyle\frac{1}{2}\sin 2x \cos 2 x \cos 4 x \cos 8 x=\displaystyle\frac{1}{16}\\ \Leftrightarrow\ &\displaystyle\frac{1}{4}\sin4x\cos4x\cos8x=\displaystyle\frac{1}{16}\\ \Leftrightarrow\ &\displaystyle\frac{1}{8}\sin8x\cos8x=\displaystyle\frac{1}{16}\\ \Leftrightarrow\ &\displaystyle\frac{1}{16}\sin16x=\displaystyle\frac{1}{16}\\ \Leftrightarrow\ &\sin16x=1\\ \Leftrightarrow\ &16x=\displaystyle\frac{\pi}{2}+k2\pi\\ \Leftrightarrow\ &x=\displaystyle\frac{\pi}{32}+k\displaystyle\frac{\pi}{8}\,(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace \displaystyle\frac{\pi}{32}+k\displaystyle\frac{\pi}{8}\ | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(2 \sin x-\sqrt{2} \sin 2 x=0\).
\(\begin{aligned}&2 \sin x-\sqrt{2} \sin 2 x=0\\ \Leftrightarrow\ &2\sin x-2\sqrt{2}\sin x\cos x=0\\ \Leftrightarrow\ &2\sin x\left( 1-\sqrt{2}\cos x\right) =0\\ \Leftrightarrow\ &\left[\begin{aligned}&\sin x=0\\&1-\sqrt{2}\cos x=0\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=k\pi\\&\cos x=\displaystyle\frac{\sqrt{2}}{2}\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=k\pi\\&\cos x=\cos\displaystyle\frac{\pi}{4}\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=k\pi\\& x=\pm\displaystyle\frac{\pi}{4}+k2\pi\end{aligned}\right.(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace x=k\pi;\pm\displaystyle\frac{\pi}{4}+k2\pi\ | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\sin 4 x+\sqrt{3} \sin 2 x=0\).
\(\begin{aligned}&\sin 4 x+\sqrt{3} \sin 2 x=0\\ \Leftrightarrow\ &2\sin 2x\cos 2x+\sqrt{3}\sin 2x=0\\ \Leftrightarrow\ &\sin 2x\left( 2\cos 2x+\sqrt{3}\right) =0\\ \Leftrightarrow\ &\left[\begin{aligned}&\sin 2x=0\\&2\cos 2x+\sqrt{3}=0\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&2x=k\pi\\&\cos 2x=-\displaystyle\frac{\sqrt{3}}{2}\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=k\displaystyle\frac{\pi}{2}\\&\cos 2x=\cos\displaystyle\frac{5\pi}{6}\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=k\displaystyle\frac{\pi}{2}\\& 2x=\pm\displaystyle\frac{5\pi}{6}+k2\pi\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=k\displaystyle\frac{\pi}{2}\\& x=\pm\displaystyle\frac{5\pi}{12}+k\pi\end{aligned}\right.(k\in\mathbb{Z}).\end{aligned}\)
Vậy \(S=\left\lbrace k\displaystyle\frac{\pi}{2};\pm\displaystyle\frac{5\pi}{12}+k\pi\ | \,k\in\mathbb{Z}\right\rbrace \).
Giải phương trình: \(\sin x-\sqrt{3} \cos \displaystyle\frac{x}{2}=0\).
Phương trình đã cho tương đương
\(\begin{aligned}&2\sin\displaystyle\frac{x}{2}\cos\displaystyle\frac{x}{2} - \sqrt{3}\cos\displaystyle\frac{x}{2} = 0 \\ \Leftrightarrow\ &\cos\displaystyle\frac{x}{2} \left( 2\sin\displaystyle\frac{x}{2} - \sqrt{3} \right)\\ \Leftrightarrow\ &\left[\begin{aligned}&\cos\displaystyle\frac{x}{2} = 0\\ &\sin\displaystyle\frac{x}{2} = \displaystyle\frac{\sqrt{3}}{2} = \sin\displaystyle\frac{\pi}{3}.\end{aligned}\right.\end{aligned}\)
+) \(\cos\displaystyle\frac{x}{2} = 0\) \(\Leftrightarrow \displaystyle\frac{x}{2} = \displaystyle\frac{\pi}{2}+k\pi\) \(\Leftrightarrow x=(2k+1)\pi\) với \(k\in\mathbb{Z}\).
+) \(\sin\displaystyle\frac{x}{2} = \sin\displaystyle\frac{\pi}{3}\) \(\Leftrightarrow \left[\begin{aligned}&\displaystyle\frac{x}{2} = \displaystyle\frac{\pi}{3} + k2\pi\\ &\displaystyle\frac{x}{2} = \displaystyle\frac{2\pi}{3} + k2\pi\end{aligned}\right.\) \(\Leftrightarrow \left[\begin{aligned}&x=\displaystyle\frac{2\pi}{3} + k4\pi\\ &x = \displaystyle\frac{4\pi}{3} + k4\pi\end{aligned}\right.\) với \(k\in\mathbb{Z}\).
Vậy \(x = (2k+1)\pi\), \(x=\displaystyle\frac{2\pi}{3} + k4\pi\) và \(x = \displaystyle\frac{4\pi}{3} + k4\pi\), với \(k\in\mathbb{Z}\).
Giải phương trình: \(\sin x+\sqrt{3} \sin \displaystyle\frac{x}{2}=0\).
Phương trình đã cho tương đương
\(\begin{aligned}&2\sin\displaystyle\frac{x}{2}\cos\displaystyle\frac{x}{2} + \sqrt{3}\sin\displaystyle\frac{x}{2} = 0 \\ \Leftrightarrow\ &\sin\displaystyle\frac{x}{2} \left( 2\cos\displaystyle\frac{x}{2} + \sqrt{3} \right)\\ \Leftrightarrow\ &\left[\begin{aligned}&\sin\displaystyle\frac{x}{2} = 0\\ &\cos\displaystyle\frac{x}{2} = \displaystyle\frac{-\sqrt{3}}{2} = \cos\displaystyle\frac{5\pi}{6}.\end{aligned}\right.\end{aligned}\)
+) \(\sin\displaystyle\frac{x}{2} = 0\) \(\Leftrightarrow \displaystyle\frac{x}{2} = k\pi\) \(\Leftrightarrow x=2k\pi\), với \(k\in\mathbb{Z}\).
+) \(\cos\displaystyle\frac{x}{2} = \cos\displaystyle\frac{5\pi}{6}\) \(\Leftrightarrow \left[\begin{aligned}&\displaystyle\frac{x}{2} = \displaystyle\frac{5\pi}{6} + k2\pi\\ &\displaystyle\frac{x}{2} = \displaystyle\frac{-5\pi}{6} + k2\pi\end{aligned}\right.\) \(\Leftrightarrow \left[\begin{aligned}&x=\displaystyle\frac{5\pi}{3} + k4\pi\\ &x = -\displaystyle\frac{5\pi}{3} + k4\pi\end{aligned}\right.\), với \(k\in\mathbb{Z}\).
Vậy \(x = 2k\pi\), \(x=\displaystyle\frac{5\pi}{3} + k4\pi\) và \(x = -\displaystyle\frac{5\pi}{3} + k4\pi\), với \(k\in\mathbb{Z}\).
Giải phương trình: \(\sin 3 x-\cos 5 x=0\).
\(pt\Leftrightarrow \sin 3x = \sin\left( \displaystyle\frac{\pi}{2} -5x \right) \Leftrightarrow \left[\begin{aligned}&3x = \displaystyle\frac{\pi}{2} - 5x + k2\pi\\ &3x = \pi - \left( \displaystyle\frac{\pi}{2} - 5x \right) + k2\pi\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&x = \displaystyle\frac{\pi}{16} + \displaystyle\frac{k\pi}{4}\\ &x = -\displaystyle\frac{\pi}{4} - k\pi\end{aligned}\right.\), với \(k\in\mathbb{Z}\).
Vậy \(x = \displaystyle\frac{\pi}{16} + \displaystyle\frac{k\pi}{4}\) và \(x = -\displaystyle\frac{\pi}{4} - k\pi\), với \(k\in\mathbb{Z}\).
Giải phương trình: \(\sin x-\cos 2 x=0\).
\(pt\Leftrightarrow \sin x = \sin\left( \displaystyle\frac{\pi}{2} -2x \right) \Leftrightarrow \left[\begin{aligned}&x = \displaystyle\frac{\pi}{2} - 2x + k2\pi\\ &x = \pi - \left( \displaystyle\frac{\pi}{2} - 2x \right) + k2\pi\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&x = \displaystyle\frac{\pi}{6} + \displaystyle\frac{k2\pi}{3}\\ &x = -\displaystyle\frac{\pi}{2} - k2\pi\end{aligned}\right.\), với \(k\in\mathbb{Z}\).
Vậy \(x = \displaystyle\frac{\pi}{6} + \displaystyle\frac{k2\pi}{3}\) và \(x = -\displaystyle\frac{\pi}{2} - k2\pi\), với \(k\in\mathbb{Z}\).
Giải phương trình: \(\tan \left(3 x-\displaystyle\frac{\pi}{5}\right)=\cot x\).
ĐKXĐ: \(\begin{cases}\cos \left(3 x-\displaystyle\frac{\pi}{5}\right)\ne 0 \\ \sin x\ne 0\end{cases} \Leftrightarrow \begin{cases}3x-\displaystyle\frac{\pi}{5}\ne \displaystyle\frac{\pi}{2} + k\pi\\ x\ne k\pi\end{cases} \Leftrightarrow \begin{cases}x\ne \displaystyle\frac{7\pi}{30} + \displaystyle\frac{k\pi}{3}\\ x\ne k\pi.\end{cases} \)
\(pt\Leftrightarrow \tan\left(3 x-\displaystyle\frac{\pi}{5}\right) = \tan\left( \displaystyle\frac{\pi}{2} - x \right) \Leftrightarrow 3x-\displaystyle\frac{\pi}{5} = \displaystyle\frac{\pi}{2} - x +k\pi \Leftrightarrow x = \displaystyle\frac{7\pi}{40} + \displaystyle\frac{k\pi}{4}\), với \(k\in\mathbb{Z}\).
Đối chiếu điều kiện ta thấy thỏa mãn
Vậy \(x = \displaystyle\frac{7\pi}{40} + \displaystyle\frac{k\pi}{4}\), với \(k\in\mathbb{Z}\).
Giải phương trình: \(\cot \left(2 x-\displaystyle\frac{3 \pi}{4}\right)=\tan \left(x-\displaystyle\frac{\pi}{6}\right)\).
ĐKXĐ: \(\begin{cases}\sin \left(2 x-\displaystyle\frac{3 \pi}{4}\right)\ne 0\\ \cos \left(x-\displaystyle\frac{\pi}{6}\right)\ne 0\end{cases} \Leftrightarrow \begin{cases}2x-\displaystyle\frac{3\pi}{4} \ne k\pi\\ x-\displaystyle\frac{\pi}{6} \ne \displaystyle\frac{\pi}{2} + k\pi\end{cases} \Leftrightarrow \begin{cases}x\ne \displaystyle\frac{3\pi}{8} + \displaystyle\frac{k\pi}{2}\\ x\ne \displaystyle\frac{2\pi}{3} + k\pi.\end{cases}\)
\(pt\Leftrightarrow \cot \left(2 x-\displaystyle\frac{3 \pi}{4}\right)=\cot \left(\displaystyle\frac{\pi}{2}+\displaystyle\frac{\pi}{6}-x\right) \Leftrightarrow 2x - \displaystyle\frac{3\pi}{4} = \displaystyle\frac{\pi}{2}+\displaystyle\frac{\pi}{6}-x + k\pi\).
Khi đó: \(x = \displaystyle\frac{17\pi}{36} + \displaystyle\frac{k\pi}{3}\), với \(k\in\mathbb{Z}\). Đối chiếu ĐKXĐ ta thấy thỏa mãn.
Vậy \(x = \displaystyle\frac{17\pi}{36} + \displaystyle\frac{k\pi}{3}\), với \(k\in\mathbb{Z}\).
Giải phương trình: \(\tan 3 x \cdot \tan x=1\).
ĐKXĐ: \(\begin{cases}\cos 3x\ne 0\\ \cos x \ne 0\end{cases} \Leftrightarrow \begin{cases}3x\ne \displaystyle\frac{\pi}{2} + k\pi\\ &x\ne \displaystyle\frac{\pi}{2} + k\pi\end{cases} \Leftrightarrow \begin{cases}x\ne \displaystyle\frac{\pi}{6} + \displaystyle\frac{k\pi}{3}\\ x\ne\displaystyle\frac{\pi}{2} + k\pi\end{cases}\).
\(pt\Leftrightarrow \tan 3x=\displaystyle\frac{1}{\tan x} = \cot x \Leftrightarrow \tan 3x = \tan\left( \displaystyle\frac{\pi}{2} - x \right) \Leftrightarrow 3x = \displaystyle\frac{\pi}{2} - x + k\pi\).
Khi đó: \(x = \displaystyle\frac{\pi}{8} + \displaystyle\frac{k\pi}{4}\), với \(k\in\mathbb{Z}\). Đối chiếu ĐKXĐ ta thấy thỏa mãn.
Vậy \(x = \displaystyle\frac{\pi}{8} + \displaystyle\frac{k\pi}{4}\), với \(k\in\mathbb{Z}\).
Giải phương trình: \(\sin (2 x+1)+\cos (3 x-1)=0\).
\(pt \Leftrightarrow \sin(2x+1) = -\cos(3x-1) = \sin \left( 3x - 1 - \displaystyle\frac{\pi}{2}\right) \Leftrightarrow \left[\begin{aligned}&2x+1 = 3x - 1 - \displaystyle\frac{\pi}{2} + k2\pi\\ &2x+1 = \pi - \left( 3x - 1 - \displaystyle\frac{\pi}{2} \right) + k2\pi.\end{aligned}\right.\)
Khi đó phương trình có nghiệm \(x = 2+\displaystyle\frac{\pi}{2} - k2\pi\) hoặc \(x = \displaystyle\frac{3\pi}{10} + \displaystyle\frac{k2\pi}{5}\) với \(k\in\mathbb{Z}\).
Giải phương trình: \(\sin \left(2 x-\displaystyle\frac{\pi}{6}\right)=-\sin \left(x-\displaystyle\frac{\pi}{4}\right)\).
\(pt\Leftrightarrow \sin \left(2 x-\displaystyle\frac{\pi}{6}\right)=\sin \left(\displaystyle\frac{\pi}{4}-x\right) \Leftrightarrow \left[\begin{aligned}&2x - \displaystyle\frac{\pi}{6} = \displaystyle\frac{\pi}{4} - x + k2\pi\\ &2x - \displaystyle\frac{\pi}{6} = \pi - \left( \displaystyle\frac{\pi}{4} - x \right) + k2\pi\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&x = \displaystyle\frac{5\pi}{36} + \displaystyle\frac{k2\pi}{3}\\ &x = \displaystyle\frac{11\pi}{12} + k2\pi.\end{aligned}\right.\)
Giải phương trình: \(\cos \left(2 x+\displaystyle\frac{\pi}{3}\right)=-\cos \left(x+\displaystyle\frac{\pi}{4}\right)\).
\(pt\Leftrightarrow \cos \left(2 x+\displaystyle\frac{\pi}{3}\right)=\cos \left(\displaystyle\frac{3\pi}{4} - x\right) \Leftrightarrow \left[\begin{aligned}&2 x+\displaystyle\frac{\pi}{3} = \displaystyle\frac{3\pi}{4} - x + k2\pi\\ &2 x+\displaystyle\frac{\pi}{3} = -\displaystyle\frac{3\pi}{4} + x + k2\pi\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}& x = \displaystyle\frac{5\pi}{36} + \displaystyle\frac{k2\pi}{3}\\ & x = -\displaystyle\frac{13\pi}{12} + k2\pi.\end{aligned}\right.\)
Giải phương trình: \(\tan \left(3 x-\displaystyle\frac{\pi}{3}\right)=-\tan x\).
ĐKXĐ: \(\begin{cases}\cos \left( 3x - \displaystyle\frac{\pi}{3} \right)\ne 0\\ \cos x \ne 0\end{cases} \Leftrightarrow \begin{cases}3x - \displaystyle\frac{\pi}{3}\ne \displaystyle\frac{\pi}{2} + k\pi\\ x\ne \displaystyle\frac{\pi}{2} + k\pi\end{cases} \Leftrightarrow \begin{cases}x\ne \displaystyle\frac{5\pi}{18} + \displaystyle\frac{k\pi}{3}\\ x\ne\displaystyle\frac{\pi}{2} + k\pi.\end{cases}\)
\(pt\Leftrightarrow \tan \left(3 x-\displaystyle\frac{\pi}{3}\right) = \tan (-x) \Leftrightarrow 3x -\displaystyle\frac{\pi}{3} = - x + k\pi \Leftrightarrow x = \displaystyle\frac{\pi}{12} + \displaystyle\frac{k\pi}{4}\), với \(k\in\mathbb{Z}\).
Đối chiếu ĐKXĐ ta thấy thỏa mãn.
Vậy \(x = \displaystyle\frac{\pi}{12} + \displaystyle\frac{k\pi}{4}\), với \(k\in\mathbb{Z}\).
Giải phương trình: \(\cot \left(x-\displaystyle\frac{\pi}{4}\right)=-\cot x\).
ĐKXĐ: \(\begin{cases}\sin \left( x-\displaystyle\frac{ \pi}{4}\right)\ne 0\\ \sin x\ne 0\end{cases} \Leftrightarrow \begin{cases}x-\displaystyle\frac{\pi}{4} \ne k\pi\\ x \ne k\pi\end{cases} \Leftrightarrow \begin{cases}x\ne \displaystyle\frac{\pi}{4} + k\pi\\ &x\ne k\pi.\end{cases}\)
\(pt\Leftrightarrow \cot \left( x-\displaystyle\frac{ \pi}{4}\right)=\cot(-x) \Leftrightarrow x - \displaystyle\frac{\pi}{4} = -x + k\pi\Leftrightarrow x = \displaystyle\frac{\pi}{8} + \displaystyle\frac{k\pi}{2}\), với \(k\in\mathbb{Z}\).
Đối chiếu ĐKXĐ ta thấy thỏa mãn.
Vậy \(x = \displaystyle\frac{\pi}{8} + \displaystyle\frac{k\pi}{2}\), với \(k\in\mathbb{Z}\).
Giải phương trình: \(\sin \left(3 x+\displaystyle\frac{2 \pi}{3}\right)+\sin \left(x-\displaystyle\frac{7 \pi}{5}\right)=0\).
\(pt\Leftrightarrow \sin \left(3 x+\displaystyle\frac{2 \pi}{3}\right)=\sin \left(-x+\displaystyle\frac{7 \pi}{5}\right) \Leftrightarrow \left[\begin{aligned}& 3 x+\displaystyle\frac{2 \pi}{3} = -x+\displaystyle\frac{7 \pi}{5} + k2\pi\\ &3 x+\displaystyle\frac{2 \pi}{3} = \pi - \left( -x+\displaystyle\frac{7 \pi}{5} \right) + k2\pi.\end{aligned}\right.\)
Khi đó phương trình có nghiệm \(x = \displaystyle\frac{11\pi}{60} + \displaystyle\frac{k\pi}{2}\) hoặc \(x = -\displaystyle\frac{31\pi}{30} + k\pi\), với \(k\in\mathbb{Z}\).
Giải phương trình: \(\cos \left(4 x+\displaystyle\frac{\pi}{3}\right)+\sin \left(x-\displaystyle\frac{\pi}{4}\right)=0\).
\(pt\Leftrightarrow \cos \left(4 x+\displaystyle\frac{\pi}{3}\right)=\sin \left(\displaystyle\frac{\pi}{4} - x\right) \Leftrightarrow \cos \left(4 x+\displaystyle\frac{\pi}{3}\right)=\cos \left(\displaystyle\frac{\pi}{4} + x\right) \Leftrightarrow \left[\begin{aligned}&4x+\displaystyle\frac{4\pi}{3} = \displaystyle\frac{\pi}{4} + x + k2\pi\\ &4x+\displaystyle\frac{4\pi}{3} = -\displaystyle\frac{\pi}{4} - x + k2\pi.\end{aligned}\right.\)
Khi đó phương trình có nghiệm \(x = -\displaystyle\frac{13\pi}{36} + \displaystyle\frac{k2\pi}{3}\) hoặc \(x = -\displaystyle\frac{19\pi}{60} + \displaystyle\frac{k2\pi}{5}\), với \(k\in\mathbb{Z}\).
Giải phương trình: \(\sin x=-\displaystyle\frac{\sqrt{3}}{2}\) với \(x \in(0 ; 2 \pi)\).
\(pt \Leftrightarrow \sin x = \sin \left(-\displaystyle\frac{\pi}{3}\right) \Leftrightarrow \left[\begin{aligned}&x = -\displaystyle\frac{\pi}{3} + k2\pi\\ & x = \pi+\displaystyle\frac{\pi}{3} + k2\pi\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&x = -\displaystyle\frac{\pi}{3} + k2\pi\\ &x = \displaystyle\frac{4\pi}{3} + k2\pi\end{aligned}\right.\), với \(k\in \mathbb{Z}\).
Do \(x \in(0 ; 2 \pi)\) nên tập nghiệm của phương trình là \(S=\left\lbrace \displaystyle\frac{4\pi}{3};\displaystyle\frac{5\pi}{3} \right\rbrace\).
Giải phương trình: \(2 \cos \left(x-\displaystyle\frac{\pi}{3}\right)=1\) với \(-\pi<x<\pi\).
\(pt \Leftrightarrow \cos \left(x-\displaystyle\frac{\pi}{3}\right)= \cos \displaystyle\frac{\pi}{3} \Leftrightarrow \left[\begin{aligned}& x - \displaystyle\frac{\pi}{3} = \displaystyle\frac{\pi}{3} + k2\pi\\ &x - \displaystyle\frac{\pi}{3} = -\displaystyle\frac{\pi}{3} + k2\pi\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&x = \displaystyle\frac{2\pi}{3} + k2\pi\\ &x = k2\pi\end{aligned}\right.\), với \(k\in \mathbb{Z}\).
Do \(-\pi<x<\pi \) nên tập nghiệm của phương trình là \(S=\left\lbrace \displaystyle\frac{2\pi}{3};0 \right\rbrace\).
Giải phương trình: \(2 \sin 2 x+1=0\) với \(0<x<90^{\circ}\).
\(pt \Leftrightarrow \sin 2x = -\displaystyle\frac{1}{2} = \sin (-30^\circ) \Leftrightarrow \left[\begin{aligned}& 2x = -30^\circ + k360^\circ\\ & 2x = 360^\circ + 30^\circ + k360^\circ\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}& x = -15^\circ + k180^\circ\\ & x = 195^\circ + k180^\circ\end{aligned}\right.\), với \(k\in \mathbb{Z}\).
Do \(0<x<90^{\circ}\) nên tập nghiệm của phương trình là \(S=\left\lbrace 15^\circ \right\rbrace\).
Giải phương trình: \(\cos ^{3} x-2 \cos ^{2} x=0\), \(\forall x \in\left[0 ; 720^{\circ}\right]\)
\(pt\Leftrightarrow \left[\begin{aligned}&\cos x = 2\ \text{(loại)}\\ & \cos x = 0\end{aligned}\right. \Leftrightarrow x = 90^\circ + k180^\circ\) với \(k\in\mathbb{Z}\).
Do \(x \in\left[0 ; 720^{\circ}\right]\) nên tập nghiệm của phương trình là
\(S=\left\lbrace 90^\circ;270^\circ;450^\circ;630^\circ \right\rbrace\).
Giải phương trình: \(\sin (\pi \sin 2 x)=1\).
\(pt\Leftrightarrow \pi\sin 2x = \displaystyle\frac{\pi}{2} + k2\pi \Leftrightarrow \sin 2x = \displaystyle\frac{1}{2} + 2k\), với \(k\in \mathbb{Z}\).
Do \(-1\le \sin 2x \le 1\) nên \(-\displaystyle\frac{3}{4} \le k\le \displaystyle\frac{1}{4}\), mà \(k\in\mathbb{Z}\) nên \(k=0\).
Khi đó \(\sin 2x = \displaystyle\frac{1}{2} = \sin \displaystyle\frac{\pi}{6} \Leftrightarrow \left[\begin{aligned}&2x = \displaystyle\frac{\pi}{6} + m2\pi\\ &2x =\pi - \displaystyle\frac{\pi}{6}+m2\pi\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&x = \displaystyle\frac{\pi}{12} + m\pi\\ &x =\displaystyle\frac{5\pi}{12}+m\pi\end{aligned}\right.\), với \(m\in\mathbb{Z}\).
Vậy phương trình có nghiệm \(x = \displaystyle\frac{\pi}{12} + m\pi\) và \(x =\displaystyle\frac{5\pi}{12}+m\pi\), với \(m\in\mathbb{Z}\).
Giải phương trình: \(\sin (\pi \cos 2 x)=1\).
\(pt\Leftrightarrow \pi\cos 2x = \displaystyle\frac{\pi}{2} + k2\pi \Leftrightarrow \cos 2x = \displaystyle\frac{1}{2} + 2k\), với \(k\in \mathbb{Z}\).
Do \(-1\le \cos 2x \le 1\) nên \(-\displaystyle\frac{3}{4} \le k\le \displaystyle\frac{1}{4}\), mà \(k\in\mathbb{Z}\) nên \(k=0\).
Khi đó
\(\cos 2x = \displaystyle\frac{1}{2} = \cos \displaystyle\frac{\pi}{3} \Leftrightarrow \left[\begin{aligned}&2x = \displaystyle\frac{\pi}{3} + m2\pi\\ &2x = - \displaystyle\frac{\pi}{3}+m2\pi\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&x = \displaystyle\frac{\pi}{6} + m\pi\\ &x =-\displaystyle\frac{\pi}{6}+m\pi\end{aligned}\right.\), với \(m\in\mathbb{Z}\).
Vậy phương trình có nghiệm \(x = \displaystyle\frac{\pi}{6} + m\pi\) và \(x =-\displaystyle\frac{\pi}{6}+m\pi\), với \(m\in\mathbb{Z}\).
Giải phương trình: \(\sin 4 x \cdot\left(2+\tan ^{2} x\right)=0\).
ĐKXĐ: \(\cos x \ne 0\Leftrightarrow x\ne \displaystyle\frac{\pi}{2} + k\pi\).
Khi đó
\(pt\Leftrightarrow \sin 4x =0 \Leftrightarrow 4x = k\pi \Leftrightarrow x =\displaystyle\frac{k\pi}{4}\), với \(k\in\mathbb{Z}\).
Đối chiếu ĐKXĐ thì tập nghiệm của phương trình là
\(S=\left\lbrace \displaystyle\frac{m\pi}{4} + k2\pi \ | \ m\in\{0;1;3;4;5;7\}, k\in\mathbb{Z} \right\rbrace\).
Giải phương trình: \((\cos 4 x-1)\left(1+\cot ^{2} x\right)=0\).
ĐKXĐ: \(\sin x \ne 0\Leftrightarrow x\ne k\pi\).
Khi đó
\(pt\Leftrightarrow \cos 4x =1 \Leftrightarrow 4x = k2\pi \Leftrightarrow x =\displaystyle\frac{k\pi}{2}\), với \(k\in\mathbb{Z}\).
Đối chiếu ĐKXĐ thì tập nghiệm của phương trình là
\(S=\left\lbrace \displaystyle\frac{m\pi}{2} + k2\pi \ | \ m\in\{1;3\}, k\in\mathbb{Z} \right\rbrace\).
Giải phương trình: \(\displaystyle\frac{\cos 2 x-1}{1-\cos x}=0\).
ĐKXĐ: \(\cos x \ne 1\Leftrightarrow x\ne k2\pi\).
Khi đó
\(pt\Leftrightarrow \cos 2x =1 \Leftrightarrow 2x = k2\pi \Leftrightarrow x =k\pi\), với \(k\in\mathbb{Z}\).
Đối chiếu ĐKXĐ thì tập nghiệm của phương trình là
\(S=\left\lbrace (2k+1)\pi \ | \ k\in\mathbb{Z} \right\rbrace\).
Giải phương trình: \(\displaystyle\frac{\cos 2 x}{\tan x-1}=0\).
ĐKXĐ:
\(\begin{cases}\tan x \ne 1\\ \cos x \ne 0\end{cases} \Leftrightarrow\begin{cases}x\ne \displaystyle\frac{\pi}{4} + k\pi\\ x\ne \displaystyle\frac{\pi}{2} + k\pi.\end{cases}\)
Khi đó
\(pt\Leftrightarrow \cos 2x =0 \Leftrightarrow 2x = \displaystyle\frac{\pi}{2} k\pi \Leftrightarrow x =\displaystyle\frac{\pi}{4} + \displaystyle\frac{k\pi}{2}\), với \(k\in\mathbb{Z}\).
Đối chiếu ĐKXĐ thì tập nghiệm của phương trình là
\(S=\left\lbrace \displaystyle\frac{m\pi}{4} + k2\pi \ | \ m\in\{3;7\}, k\in\mathbb{Z} \right\rbrace\).
Giải phương trình: \(\sin 5 x+\sin 3 x+\sin x=0\).
\(\begin{align*}pt&\Leftrightarrow \sin 3x + 2\sin 3x \cos 2x = 0 \Leftrightarrow \sin 3x (1+2\cos 2x) = 0 \\ &\Leftrightarrow \left[\begin{aligned}&\sin 3x = 0\\ &\cos 2x = -\displaystyle\frac{1}{2} = \cos\displaystyle\frac{2\pi}{3}\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&3x =k\pi\\ &2x = \displaystyle\frac{2\pi}{3} + k2\pi\\ &2x=-\displaystyle\frac{2\pi}{3}+k2\pi.\end{aligned}\right.\end{align*}\)
Khi đó phương trình có nghiệm \(x=\displaystyle\frac{k\pi}{3}\), \(x=\displaystyle\frac{\pi}{3} + k\pi\) và \(x=-\displaystyle\frac{\pi}{3} + k\pi\) với \(k\in\mathbb{Z}\).
Giải phương trình: \(\sin 7 x+\sin 4 x+\sin x=0\).
\begin{align*}pt&\Leftrightarrow \sin 4x + 2\sin 4x \cos 3x = 0\Leftrightarrow \sin 4x (1+2\cos 3x) = 0 \\&\Leftrightarrow \left[\begin{aligned}&\sin 4x = 0\\ &\cos 3x = -\displaystyle\frac{1}{2} = \cos\displaystyle\frac{2\pi}{3}\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&4x =k\pi\\ &3x = \displaystyle\frac{2\pi}{3} + k2\pi\\ &3x=-\displaystyle\frac{2\pi}{3}+k2\pi.\end{aligned}\right.\end{align*}
Khi đó phương trình có nghiệm \(x=\displaystyle\frac{k\pi}{4}\) hoặc \(x=\displaystyle\frac{2\pi}{9} + \displaystyle\frac{k\pi}{3}\) hoặc \(x=-\displaystyle\frac{2\pi}{9} + \displaystyle\frac{k\pi}{3}\) với \(k\in\mathbb{Z}\).
Giải phương trình: \(\cos x+\cos 3 x+\cos 5 x=0\)
\begin{align*}pt&\Leftrightarrow \cos 3x + 2\cos 3x \cos 2x = 0\Leftrightarrow \cos 3x(2\cos 2x + 1) = 0\\&\Leftrightarrow \left[\begin{aligned}&\cos 3x = 0\\ &\cos 2x = -\displaystyle\frac{1}{2} = \cos\displaystyle\frac{2\pi}{3}\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&3x =\displaystyle\frac{\pi}{2}+k\pi\\ &2x = \displaystyle\frac{2\pi}{3} + k2\pi\\ &2x=-\displaystyle\frac{2\pi}{3}+k2\pi.\end{aligned}\right.\end{align*}
Khi đó phương trình có nghiệm \(x=\displaystyle\frac{\pi}{6}+\displaystyle\frac{k\pi}{3}\), \(x=\displaystyle\frac{\pi}{3} + k\pi\) và \(x=-\displaystyle\frac{\pi}{3} + k\pi\) với \(k\in\mathbb{Z}\).
Giải phương trình: \(\sin x-4 \cos x+\sin 3 x=0\).
\(pt\Leftrightarrow 2\sin 2x \cos x - 4\cos x= 0\Leftrightarrow \cos x(\sin 2x - 2) = 0 \Leftrightarrow \left[\begin{aligned}&\cos x = 0 \Leftrightarrow x =\displaystyle\frac{\pi}{2}+k\pi\\ &\sin 2x = 2\ (\text{loại}).\end{aligned}\right.\)
Vậy phương trình có nghiệm \(x=\displaystyle\frac{\pi}{2} + k\pi\) với \(k\in\mathbb{Z}\).
Giải phương trình: \(\sin 3 x+\cos 2 x-\sin x=0\).
\begin{align*}pt &\Leftrightarrow \cos 2x + 2\sin x \cos 2x = 0\Leftrightarrow \cos 2x(1+2\sin x) = 0\\&\Leftrightarrow \left[\begin{aligned}&\cos 2x = 0\\ &\sin x = -\displaystyle\frac{1}{2} = \sin\left( -\displaystyle\frac{\pi}{6} \right)\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&2x =\displaystyle\frac{\pi}{2}+k\pi\\ &x = -\displaystyle\frac{\pi}{6} + k2\pi\\ &x=\pi+\displaystyle\frac{\pi}{6}+k2\pi.\end{aligned}\right.\end{align*}
Vậy phương trình có nghiệm \(x=\displaystyle\frac{\pi}{4}+\displaystyle\frac{k\pi}{2}\), \(x=-\displaystyle\frac{\pi}{6} + k\pi\) và \(x=\displaystyle\frac{7\pi}{6} + k\pi\) với \(k\in\mathbb{Z}\).
Giải phương trình: \(\cos 3 x+2 \sin 2 x-\cos x=0\).
\(pt\Leftrightarrow 2\sin 2x - 2\sin 2x \sin x = 0\Leftrightarrow \sin 2x(1-\sin x) = 0 \Leftrightarrow \left[\begin{aligned}&\sin 2x = 0\\ &\sin x = 1\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&2x =k\pi\\ &x=\displaystyle\frac{\pi}{2}+k2\pi.\end{aligned}\right.\)
Vậy phương trình có nghiệm \(x=\displaystyle\frac{k\pi}{2}\) và \(x=\displaystyle\frac{\pi}{2} + k\pi\) với \(k\in\mathbb{Z}\).
Giải phương trình: \(\cos 3 x+\cos 2 x+\cos x+1=0\).
\(\begin{aligned}& \cos 3 x+\cos 2 x+\cos x+1=0\\ \Leftrightarrow\ & (\cos 3 x+\cos x)+(1+\cos 2 x)=0\\ \Leftrightarrow\ & 2 \cos 2 x \cdot \cos x+2 \cos ^2 x=0\\ \Leftrightarrow\ & 2\cos x\cdot(\cos 2x +\cos x)=0\\ \Leftrightarrow\ & 4\cos x\cdot \cos \displaystyle\frac{3x}{2}\cdot \cos \displaystyle\frac{x}{2} =0\\ \Leftrightarrow\ & \left[\begin{aligned}&\cos x=0\\&\cos \displaystyle\frac{3x}{2}=0\\&\cos \displaystyle\frac{x}{2} =0\end{aligned}\right.\\ \Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{\pi}{2}+k\pi\\&\displaystyle\frac{3x}{2}=\displaystyle\frac{\pi}{2}+k\pi \\&\displaystyle\frac{x}{2}=\displaystyle\frac{\pi}{2}+k\pi \end{aligned}\right. \\ \Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{\pi}{2}+k\pi \\&x=\displaystyle\frac{\pi}{3}+\displaystyle\frac{k2\pi}{3}\end{aligned}\right.\,\,(\text{với } k\in \mathbb{Z}).\end{aligned}\)
Giải phương trình: \(1-\sin x-\cos 2 x+\sin 3 x=0\).
\(\begin{aligned}& 1-\sin x-\cos 2 x+\sin 3 x=0\\ \Leftrightarrow\ & (1-\cos 2x) +(\sin 3 x- \sin x)=0 \\ \Leftrightarrow\ & 2\sin^2x +2\cos 2x\sin x=0\\ \Leftrightarrow\ & 2\sin x\cdot(\sin x+\cos 2x)=0 \\ \Leftrightarrow\ & 2\sin x\cdot(-2\sin^2x+\sin x+1)=0 \\ \Leftrightarrow\ & \left[\begin{aligned}&\sin x=0\\&\sin x=1\\&\sin x=-\displaystyle\frac{1}{2}\end{aligned}\right.\\ \Leftrightarrow\ & \left[\begin{aligned}&x=k\pi\\&x=\displaystyle\frac{\pi}{2}+k2\pi\\&x=-\displaystyle\frac{\pi}{6}+k2\pi\end{aligned}\right.\,\, (\text{với } k\in \mathbb{Z}).\end{aligned}\)
Giải phương trình: \(\sin 5 x+\sin x+2 \sin ^2 x=1\).
\(\begin{aligned}& \sin 5 x+\sin x+2 \sin ^2 x=1\\ \Leftrightarrow\ & \sin 5 x+\sin x-(1-2 \sin ^2)=0\\ \Leftrightarrow\ & 2\sin 3x\cos 2x-\cos 2x=0\\ \Leftrightarrow\ & \cos 2x\cdot (2\sin 3x -1)=0\\ \Leftrightarrow\ & \left[\begin{aligned}&\cos 2x=0\\&\sin 3x=\displaystyle\frac{1}{2}\end{aligned}\right.\\ \Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{\pi}{4}+\displaystyle\frac{k\pi}{2}\\&x=\displaystyle\frac{\pi}{18}+\displaystyle\frac{k2\pi}{3}\\&x=\displaystyle\frac{5\pi}{18}+\displaystyle\frac{k2\pi}{3}\end{aligned}\right.(\text{với } k\in \mathbb{Z}).\end{aligned}\)
Giải phương trình: \(\sin 3 x+\cos 2 x-\cos x=0\).
\(\begin{aligned}& \sin 3 x+\cos 2 x-\cos x=0\\ \Leftrightarrow\ &2\sin \displaystyle\frac{3x}{2}\cos \displaystyle\frac{3x}{2}-2\sin \displaystyle\frac{3x}{2}\sin \displaystyle\frac{x}{2}=0 \\ \Leftrightarrow\ & 2\sin \displaystyle\frac{3x}{2}\left(\cos \displaystyle\frac{3x}{2}-\sin \displaystyle\frac{x}{2}\right)=0 \\ \Leftrightarrow\ &\left[\begin{aligned}&\sin \displaystyle\frac{3x}{2}=0\\ &\cos \displaystyle\frac{3x}{2}=\sin \displaystyle\frac{x}{2}\end{aligned}\right. \\ \Leftrightarrow\ &\left[\begin{aligned}&\sin \displaystyle\frac{3x}{2}=0\\ &\cos \displaystyle\frac{3x}{2}=\cos \left(\displaystyle\frac{\pi}{2}- \displaystyle\frac{x}{2}\right)\end{aligned}\right. \\ \Leftrightarrow\ &\left[\begin{aligned}&x=\displaystyle\frac{k2\pi}{3} \\& x=\displaystyle\frac{\pi}{4}+k\pi \\&x=-\displaystyle\frac{\pi}{2}+k2\pi.\end{aligned}\right. \,\,(\text{với } k\in \mathbb{Z}).\end{aligned}\)
Giải phương trình: \(\cos 3 x-2 \sin 2 x-\cos x-\sin x=1\).
\(\begin{aligned}& \cos 3 x-2 \sin 2 x-\cos x-\sin x=1\\ \Leftrightarrow\ & \cos 3 x-\cos x-2 \sin 2 x-\sin x=1\\ \Leftrightarrow\ & -2\sin 2x\sin x\-2\sin -\sin x -1=0\\ \Leftrightarrow\ &-2\sin 2x (\sin x+1)-(1+\sin x)=0 \\ \Leftrightarrow\ & -(\sin x+1)(2\sin 2x +1)=0\\ \Leftrightarrow\ & \left[\begin{aligned}&\sin x+1=0 \\&2\sin 2x +1=0 \end{aligned}\right.\\ \Leftrightarrow\ & \left[\begin{aligned}&\sin x=-1 \\&\sin 2x =-\displaystyle\frac{1}{2}\end{aligned}\right.\\ \Leftrightarrow\ & \left[\begin{aligned}&x=-\displaystyle\frac{\pi}{2}+k2\pi\\&x=-\displaystyle\frac{\pi}{6}+k2\pi \\&x=\displaystyle\frac{7\pi}{6}+k2\pi\end{aligned}\right., k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(4 \sin 3 x+\sin 5 x=2 \sin x \cos 2 x\).
\(\begin{aligned}& 4 \sin 3 x+\sin 5 x=2 \sin x \cos 2 x\\ \Leftrightarrow\ & 4 \sin 3 x+\sin 5x= \sin 3x -\sin x\\ \Leftrightarrow\ & 3\sin 3x +\sin 5x +\sin x=0 \\ \Leftrightarrow\ & 3\sin 3x +2\sin 3x\sin 2x=0\\ \Leftrightarrow\ & \sin 3x \cdot(3+2\sin 2x)=0\\ \Leftrightarrow\ & \sin 3x=0\\ \Leftrightarrow\ & 3x=\displaystyle\frac{\pi}{2}+k\pi\\ \Leftrightarrow\ & x=\displaystyle\frac{\pi}{6}+\displaystyle\frac{k\pi}{3} , k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin ^2\left(2 x-\displaystyle\frac{\pi}{4}\right)=\displaystyle\frac{1}{2}\).
\(\begin{aligned}& \sin ^2\left(2 x-\displaystyle\frac{\pi}{4}\right)=\displaystyle\frac{1}{2}\\ \Leftrightarrow\ & \displaystyle\frac{1}{2}\cdot \left[1-\cos \left(4x-\displaystyle\frac{\pi}{2}\right) \right]=\displaystyle\frac{1}{2} \\\Leftrightarrow\ & \cos \left(4x-\displaystyle\frac{\pi}{2}\right)=0 \\\Leftrightarrow\ & 4x-\displaystyle\frac{\pi}{2}=\displaystyle\frac{\pi}{2}+k\pi\\ \Leftrightarrow\ & x=\displaystyle\frac{\pi}{4}+k\displaystyle\frac{\pi}{4} , k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\cos ^2 x=\displaystyle\frac{2+\sqrt{3}}{4}\).
\(\begin{aligned}& \cos ^2 x=\displaystyle\frac{2+\sqrt{3}}{4}\\\Leftrightarrow\ &\displaystyle\frac{1}{2}\cdot \left(1+ \cos 2x\right)=\displaystyle\frac{2+\sqrt{3}}{4} \\\Leftrightarrow\ & \cos 2x=\displaystyle\frac{\sqrt{3}}{2}\\\Leftrightarrow\ & x=\pm\displaystyle\frac{\pi}{6}+k\pi, k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin ^2\left(3 x+\displaystyle\frac{2 \pi}{3}\right)=\sin ^2\left(\displaystyle\frac{7 \pi}{4}-x\right)\).
\(\begin{aligned}& \sin ^2\left(3 x+\displaystyle\frac{2 \pi}{3}\right)=\sin ^2\left(\displaystyle\frac{7 \pi}{4}-x\right)\\\Leftrightarrow\ &\displaystyle\frac{1}{2}\left(1-\cos \left(6x+\displaystyle\frac{4 \pi}{3} \right) \right)= \displaystyle\frac{1}{2}\left(1-\cos \left(\displaystyle\frac{7 \pi}{2}-2x\right) \right) \\\Leftrightarrow\ & \cos \left(6x+\displaystyle\frac{4 \pi}{3} \right)=\cos \left(\displaystyle\frac{7 \pi}{2}-2x\right)\\\Leftrightarrow\ & 6x+\displaystyle\frac{4 \pi}{3}=\pm\left( \displaystyle\frac{7 \pi}{2}-2x\right)+k2\pi\\\Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{13\pi}{48}+\displaystyle\frac{k\pi}{4}\\&x=-\displaystyle\frac{29\pi}{24}+\displaystyle\frac{k\pi}{2}\end{aligned}\right., k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin ^2\left(3 x+\displaystyle\frac{2 \pi}{5}\right)-\cos ^2\left(\displaystyle\frac{x}{4}-\pi\right)=0\).
\(\begin{aligned}& \sin ^2\left(3 x+\displaystyle\frac{2 \pi}{5}\right)-\cos ^2\left(\displaystyle\frac{x}{4}-\pi\right)=0\\\Leftrightarrow\ &\displaystyle\frac{1}{2}\left(1-\cos \left(6x+\displaystyle\frac{4\pi}{5} \right) \right) -\displaystyle\frac{1}{2}\left(1+\cos \left(\displaystyle\frac{x}{2}-2\pi \right) \right)=0\\\Leftrightarrow\ & \cos \left(6x+\displaystyle\frac{4\pi}{5} \right) +\cos \left(\displaystyle\frac{x}{2}-2\pi \right)=0\\\Leftrightarrow\ & 2\cos \left(\displaystyle\frac{13x}{4}-\displaystyle\frac{3\pi}{5}\right)\cdot \cos \left(\displaystyle\frac{11x}{4}+\displaystyle\frac{7\pi}{5} \right)=0\\\Leftrightarrow\ & \left[\begin{aligned}&\cos \left(\displaystyle\frac{13x}{4}-\displaystyle\frac{3\pi}{5}\right)=0\\&\cos \left(\displaystyle\frac{11x}{4}+\displaystyle\frac{7\pi}{5} \right)=0\end{aligned}\right.\\\Leftrightarrow\ &\left[\begin{aligned}&\displaystyle\frac{13x}{4}-\displaystyle\frac{3\pi}{5}=\displaystyle\frac{\pi}{2}+k\pi \\&\displaystyle\frac{11x}{4}+\displaystyle\frac{7\pi}{5}=\displaystyle\frac{\pi}{2}+k\pi\end{aligned}\right.\\\Leftrightarrow\ &\left[\begin{aligned}&x=\displaystyle\frac{22\pi}{65}+\displaystyle\frac{4k\pi}{13}\\&x=-\displaystyle\frac{18\pi}{55}+\displaystyle\frac{4k\pi}{11}\end{aligned}\right., k\in \mathbb{Z}. \end{aligned}\)
Giải phương trình: \(\sin ^2 2 x+\sin ^2 x=1\).
\(\begin{aligned}& \sin ^2 2 x+\sin ^2 x=1\\\Leftrightarrow\ & \sin ^2 2x=1-\sin ^2 x\\\Leftrightarrow\ & \sin ^2 2x=\cos^2 x\\\Leftrightarrow\ &\displaystyle\frac{1}{2}\left(1-\cos 4x \right)= \displaystyle\frac{1}{2}\left(1+\cos 2x \right)\\\Leftrightarrow\ & \cos 4x+\cos 2x =0\\\Leftrightarrow\ & 2\cos 3x\cdot \cos x=0\\\Leftrightarrow\ & \left[\begin{aligned}&\cos 3x=0\\&\cos x=0\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&3x=\displaystyle\frac{\pi}{2}+k\pi\\& x=\displaystyle\frac{\pi}{2}+k\pi\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{\pi}{6}+\displaystyle\frac{k\pi}{3}\\&x=\displaystyle\frac{\pi}{2}+k\pi,\end{aligned}\right. k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin ^2 2 x+\cos ^2 3 x=1\).
\(\begin{aligned}& \sin ^2 2 x+\cos ^2 3 x=1\\\Leftrightarrow\ & \cos ^2 3x - (1-\sin ^2 2 x)=0\\\Leftrightarrow\ & \cos ^2 3x -\cos^2 2x=0\\\Leftrightarrow\ &\left(\cos 3x-\cos 2x \right)\left(\cos 3x+\cos 2x \right)=0 \\\Leftrightarrow\ &\left[\begin{aligned}&\cos 3x-\cos 2x=0 \\&\cos 3x+\cos 2x =0\end{aligned}\right. \\\Leftrightarrow\ &\left[\begin{aligned}&-2\sin \displaystyle\frac{5x}{2}\sin \displaystyle\frac{x}{2}=0 \\&2\cos\displaystyle\frac{5x}{2}\cos \displaystyle\frac{x}{2} =0\end{aligned}\right. \\\Leftrightarrow\ &\left[\begin{aligned}&\sin \displaystyle\frac{5x}{2}=0\\&\sin \displaystyle\frac{x}{2}=0\\&\cos\displaystyle\frac{5x}{2}=0\\&\cos \displaystyle\frac{x}{2} =0\end{aligned}\right.\\\Leftrightarrow\ &\left[\begin{aligned}&\displaystyle\frac{5x}{2}=k\pi\\&\displaystyle\frac{x}{2}=k\pi\\&\displaystyle\frac{5x}{2}=\displaystyle\frac{\pi}{2}+k\pi\\&\displaystyle\frac{x}{2}=\displaystyle\frac{\pi}{2}+k\pi\end{aligned}\right.\\\Leftrightarrow\ &\left[\begin{aligned}&x=\displaystyle\frac{k2\pi}{5}\\&x=k2\pi\\&x=\displaystyle\frac{\pi}{5}+\displaystyle\frac{k2\pi}{5}\\&x=\pi+k2\pi\end{aligned}\right.\\ \Leftrightarrow\ &\left[\begin{aligned}&x=k\pi\\&x=\displaystyle\frac{k\pi}{5}\end{aligned}\right.,\,\, k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin ^2 x+\sin ^2 2 x+\sin ^2 3 x=\displaystyle\frac{3}{2}\).
\(\begin{aligned}& \sin ^2 x+\sin ^2 2 x+\sin ^2 3x=\displaystyle\frac{3}{2}\\\Leftrightarrow\ & \left(1- 2\sin ^2 x\right)+\left(1-2\sin ^2 2x \right)+\left(1- 2\sin ^2 3x\right)=0 \\\Leftrightarrow\ & \cos 2x+ \cos 4x+\cos 6x=0\\\Leftrightarrow\ & \cos 6x+\cos 2x + \cos 4x=0\\\Leftrightarrow\ & 2\cos 4x\cos 2x + \cos 4x=0\\\Leftrightarrow\ & \cos 4x\cdot(2\cos 2x+1)=0\\\Leftrightarrow\ & \left[\begin{aligned}&\cos 4x=0\\&\cos 2x=-\displaystyle\frac{1}{2}\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&4x=\displaystyle\frac{\pi}{2}+k\pi\\& 2x=\pm \displaystyle\frac{2\pi}{3}+k2\pi\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{\pi}{8}+\displaystyle\frac{k\pi}{4}\\&x=\pm \displaystyle\frac{\pi}{3}+k\pi\end{aligned}\right.,\,\, k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin ^2 2 x-\cos ^2 8 x=\displaystyle\frac{1}{2} \cos 10 x\).
\(\begin{aligned}& \sin ^2 2 x-\cos ^2 8 x=\displaystyle\frac{1}{2} \cos 10 x\\\Leftrightarrow\ &-(1-\sin ^2 2 x)+(1-2\cos ^2 8 x)=\cos 10 x \\\Leftrightarrow\ & - \cos 4x -\cos 16x=\cos 10x\\\Leftrightarrow\ & \cos 4x +\cos 16x+\cos 10x=0\\\Leftrightarrow\ & 2\cos 10x\cos 6x +\cos 10x=0\\\Leftrightarrow\ & \cos 10x(2\cos 6x+1)=0\\\Leftrightarrow\ & \left[\begin{aligned}&\cos 10x=0\\&\cos 6x=-\displaystyle\frac{1}{2}\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&10x=\displaystyle\frac{\pi}{2}+k\pi\\&6x=\pm \displaystyle\frac{2\pi}{3}+k2\pi\end{aligned}\right.\\ \Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{\pi}{20}+\displaystyle\frac{k\pi}{10} \\&x=\pm \displaystyle\frac{\pi}{9}+\displaystyle\frac{k\pi}{3}\end{aligned}\right.,\,\, k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin ^2 x+\sin ^2 3 x=\cos ^2 2 x+\cos ^2 4 x\).
\(\begin{aligned}& \sin ^2 x+\sin ^2 3 x=\cos ^2 2 x+\cos ^2 4 x\\\Leftrightarrow\ &\displaystyle\frac{1}{2}\left(1-\cos 2x \right)+\displaystyle\frac{1}{2}\left(1-\cos 6x \right)=\displaystyle\frac{1}{2}\left(1+\cos 4x \right)+\displaystyle\frac{1}{2}\left(1+\cos 8x \right) \\\Leftrightarrow\ & \cos 8x+\cos 2x+\cos 6x +\cos 4x=0 \\\Leftrightarrow\ & 2\cos 5x\cos 3x+2\cos 5x\cos x=0\\\Leftrightarrow\ & 2\cos 5x(\cos 3x+\cos x)=0\\\Leftrightarrow\ & 4\cos 5x\cdot\cos 2x\cdot\cos x =0\\\Leftrightarrow\ & \left[\begin{aligned}&\cos 5x=0\\&\cos 2x=0\\&\cos x=0\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&5x=\displaystyle\frac{\pi}{2}+k\pi\\&2x=\displaystyle\frac{\pi}{2}+k\pi\\&x=\displaystyle\frac{\pi}{2}+k\pi\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{\pi}{10}+\displaystyle\frac{k\pi}{5}\\&x=\displaystyle\frac{\pi}{4}+\displaystyle\frac{k\pi}{2}\\&x=\displaystyle\frac{\pi}{2}+k\pi,\end{aligned}\right.\,\, k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin ^2 3 x-\cos ^2 4 x=\sin ^2 5 x-\cos ^2 6 x\).
\(\begin{aligned}&\sin ^2 3 x-\cos ^2 4 x=\sin ^2 5 x-\cos ^2 6 x\\\Leftrightarrow\ & -(1-2\sin ^2 3 x)-(2\cos ^2 4 x-1)=-(1-2\sin ^2 5 x)-(2\cos ^2 6 x-1)\\\Leftrightarrow\ & -\cos 6x-\cos 8x=-\cos 10x-\cos 12x \\\Leftrightarrow\ & (\cos 12x-\cos 6x)+(\cos 10x-\cos 8x)=0\\\Leftrightarrow\ & -2\sin 9x\sin 3x-2\sin 9x\sin x=0\\\Leftrightarrow\ & -2\sin 9x\cdot(\sin 3x+\sin x)=0\\\Leftrightarrow\ & -4\sin 9x\cdot\sin 2x\cdot\sin x=0\\\Leftrightarrow\ & \left[\begin{aligned}&\sin 9x=0\\&\sin 2x=0\\&\sin x=0\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&9x=k\pi\\&2x=k\pi\\&x=k\pi\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{k\pi}{9}\\&x=\displaystyle\frac{k\pi}{2}\\&x=k\pi\end{aligned}\right.,\,\, k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\cos ^2 3 x \cdot \cos 2 x-\cos ^2 x=0\).
\(\begin{aligned}& \cos ^2 3 x \cdot \cos 2 x-\cos ^2 x=0\\\Leftrightarrow\ &\displaystyle\frac{1}{2}(1+\cos 6x) \cdot \cos 2 x-\displaystyle\frac{1}{2}(1+\cos 2x)=0 \\\Leftrightarrow\ & \cos 6x\cos 2x-1=0\\\Leftrightarrow\ &\displaystyle\frac{1}{2}(\cos 8x+\cos 4x)-1=0 \\\Leftrightarrow\ &\cos 8x+\cos 4x-2=0 \\\Leftrightarrow\ &\cos 8x-1+\cos 4x-1=0 \\\Leftrightarrow\ & -\sin^2 4x-\sin^2 2x=0 \\\Leftrightarrow\ & \sin^2 4x+\sin^2 2x=0\\\Leftrightarrow\ & \begin{cases}\sin 4x=0\\\sin 2x =0\end{cases}\\\Leftrightarrow\ & \begin{cases}4x=k\pi\\2x =k\pi\end{cases}\\\Leftrightarrow\ & \begin{cases}x=\displaystyle\frac{k\pi}{4}\\x = \displaystyle\frac{k\pi}{2}\end{cases}\\\Leftrightarrow\ & x=\displaystyle\frac{k\pi}{2},\,\, k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\cos ^2 x+\cos ^2 2 x+\cos ^2 3 x+\cos ^2 4 x=1{,}5\).
\(\begin{aligned}& \cos ^2 x+\cos ^2 2 x+\cos ^2 3 x+\cos ^2 4 x=\displaystyle\frac{3}{2}\\\Leftrightarrow\ & 2\cos ^2 x-1+2\cos ^2 2x-1+2\cos ^2 3x-1+2\cos ^2 4x=0\\\Leftrightarrow\ &\cos 2x +\cos 4x+\cos 6x+2\cos ^2 4x=0 \\\Leftrightarrow\ & \cos 2x +\cos 6x+\cos 4x+2\cos ^2 4x=0 \\\Leftrightarrow\ & 2\cos 4x\cos 2x +\cos 4x+2\cos ^2 4x=0 \\\Leftrightarrow\ & \cos 4x\cdot(2\cos 2x+1+2\cos 4x)=0 \\\Leftrightarrow\ & \cos 4x\cdot(4\cos^2 2x+2\cos 2x-1)=0 \\\Leftrightarrow\ & \left[\begin{aligned}&\cos 4x=0\\&\cos 2x=\displaystyle\frac{-1\pm\sqrt{5}}{4}\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&4x=\displaystyle\frac{\pi}{2}+k\pi\\& 2x=\pm \arccos\left(\displaystyle\frac{-1\pm\sqrt{5}}{4} \right)+k2\pi\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{\pi}{8}+\displaystyle\frac{k\pi}{4}\\&x=\pm\displaystyle\frac{1}{2}\arccos\left(\displaystyle\frac{-1\pm\sqrt{5}}{4} \right)+k\pi\end{aligned}\right.,\,\, k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin x \sin 7 x=\sin 3 x \sin 5 x\).
\(\begin{aligned}& \sin x \sin 7 x=\sin 3 x \sin 5 x\\\Leftrightarrow\ & \sin 7 x\sin x =\sin 5 x\sin 3 x \\\Leftrightarrow\ &-\displaystyle\frac{1}{2}\left(\cos 8x-\cos 6x \right)=-\displaystyle\frac{1}{2}\left(\cos 8x-\cos 2x \right) \\\Leftrightarrow\ & \cos 6x-\cos 2x=0\\\Leftrightarrow\ & -2\sin 4x\sin 2x=0\\\Leftrightarrow\ & \left[\begin{aligned}&\sin 4x=0\\&\sin 2x=0\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{k\pi}{4}\\&x=\displaystyle\frac{k\pi}{2}\end{aligned}\right.\\\Leftrightarrow\ & x=\displaystyle\frac{k\pi}{4},\,\, k\in \mathbb{Z}. \end{aligned}\)
Giải phương trình: \(\sin x \sin 3 x+\sin 4 x \sin 8 x=0\).
\(\begin{aligned}& \sin x \sin 3 x+\sin 4 x \sin 8 x=0\\\Leftrightarrow\ & \sin 3 x\sin x +\sin 8 x\sin 4 x =0\\\Leftrightarrow\ & -\displaystyle\frac{1}{2}\left(\cos 4x-\cos 2x \right)-\displaystyle\frac{1}{2}\left(\cos 12x-\cos 4x \right)=0\\\Leftrightarrow\ & \cos 2x-\cos 12x=0\\\Leftrightarrow\ & \cos 12x-\cos 2x=0\\\Leftrightarrow\ & -2\sin 7x\sin 5x =0\\\Leftrightarrow\ & \left[\begin{aligned}&\sin 7x=0\\&\sin 5x =0\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&7x=k\pi\\&5x=k\pi\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{k\pi}{7} \\&x=\displaystyle\frac{k\pi}{5} \end{aligned}\right.,\,\,k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin 4 x \sin 2 x+\sin 9 x \sin 3 x=\cos ^{2} x\).
Phương trình đã cho tương đương
\(\begin{aligned}& \cos 2x - \cos 6x+ \cos 6x- cos 12x=2\cos^2x\\\Leftrightarrow\ & \cos 2x -\cos 12x= \cos 2x+1\\\Leftrightarrow\ & \cos 12x =-1 \Leftrightarrow 12x=\pi +k2\pi\\\Leftrightarrow\ & x=\displaystyle\frac{\pi}{12}+\displaystyle\frac{k\pi}{6}, k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(4 \sin x \sin 2 x \sin 3 x=\sin 4 x\).
Phương trình đã cho tương đương
\(\begin{aligned}& \sin 2x\left(2\sin x \sin 3x-\cos 2x\right)=0\\\Leftrightarrow\ & \sin 2x \left(\cos 2x-\cos4x -\cos 2x\right)=0\\\Leftrightarrow\ & \left[\begin{aligned}&\sin 2x=0\\ &\cos 4x=0\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&x=\displaystyle\frac{k\pi}{2}\\ &x=\displaystyle\frac{\pi}{8}+ \displaystyle\frac{k\pi}{4}\end{aligned}\right., k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\cos 8 x \cos 5 x=\cos 7 x \cos 4 x\).
Phương trình đã cho tương đương
\(\begin{aligned}& \cos 13x+\cos 3x=\cos 11x+\cos 3x\\\Leftrightarrow\ & \left[\begin{aligned}&13x=11x+k2\pi\\ &13x=-11x+k2\pi\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&x=k\pi\\ &x=\displaystyle\frac{k\pi}{12}\end{aligned}\right.\Leftrightarrow x=\displaystyle\frac{k\pi}{12}, k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\cos x \cos 5 x=\cos 2 x \cos 4 x\)
Phương trình đã cho tương đương
\(\begin{aligned}& \cos 6x+\cos 4x=\cos 6x+\cos 2x\\\Leftrightarrow\ & \left[\begin{aligned}&4x=2x+k2\pi\\&4x=-2x+k2\pi\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&x=k\pi\\ &x=\displaystyle\frac{k\pi}{3}\end{aligned}\right.\\\Leftrightarrow\ & x=\displaystyle\frac{k\pi}{3}, k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin 7 x \cos x=\sin 5 x \cos 3 x\).
Phương trình đã cho tương đương
\(\begin{aligned}& \sin 8x+\sin 6x=\sin 8x+\sin 2x\\\Leftrightarrow\ & \left[\begin{aligned}&6x=2x+k2\pi\\ &6x=\pi-2x+k2\pi\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{k\pi}{2}\\ &x=\displaystyle\frac{\pi}{8}+ \displaystyle\frac{k\pi}{4}\end{aligned}\right., k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\cos 5 x \sin 4 x=\cos 3 x \sin 2 x\).
Phương trình đã cho tương đương
\(\begin{aligned}& \sin 9x-\sin x=\sin 5x-\sin x\\\Leftrightarrow\ & \left[\begin{aligned}&9x=5x+k2\pi\\ &9x=\pi-5x+k2\pi\end{aligned}\right.\\\Leftrightarrow\ & \left[\begin{aligned}&x=\displaystyle\frac{k\pi}{2}\\ &x=\displaystyle\frac{\pi}{14}+ \displaystyle\frac{k\pi}{7}\end{aligned}\right., k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(2 \cos x+\sqrt{3} \sin x=\sin 2 x+\sqrt{3}\).
Phương trình đã cho tương đương
\(\begin{aligned}& 2\cos x-\sqrt{3}+\sqrt{3}\sin x-2\sin x\cos x=0\\\Leftrightarrow\ & \left(2\cos x-\sqrt{3}\right)\left(1-\sin x\right)=0\\\Leftrightarrow\ & \left[\begin{aligned}&\cos x=\displaystyle\frac{\sqrt{3}}{2}\\ &\sin x=1\end{aligned}\right.\\ \Leftrightarrow\ & \left[\begin{aligned}&x=\pm \displaystyle\frac{\pi}{6}+k2\pi\\ & x=\displaystyle\frac{\pi}{2}+k2\pi\end{aligned}\right., k\in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin x-4 \cos x=2-\sin 2 x\).
Phương trình đã cho tương đương
\(\begin{aligned}& \sin x-2=4\cos x-2\sin x\cos x\\\Leftrightarrow\ & \left[\begin{aligned}&\sin x-2=0\\ &2\cos x=-1\end{aligned}\right.\\\Leftrightarrow\ & x=\pm \displaystyle\frac{2\pi}{3}+k2\pi, k \in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin x+4 \cos x=2+\sin 2 x\).
Phương trình đã cho tương đương
\(\begin{aligned}& \sin x-2=2\sin x\cos x-4\cos x\\\Leftrightarrow\ & \left[\begin{aligned}&\sin x-2=0\\ &2\cos x=1\end{aligned}\right.\\\Leftrightarrow\ & x=\pm \displaystyle\frac{\pi}{3}+k2\pi, k \in \mathbb{Z}.\end{aligned}\)
Giải phương trình: \(\sin 2 x-2 \sin x-2 \cos x+2=0\).
Phương trình đã cho tương đương
\(\begin{aligned}& \sin x\cos x-\sin x-\cos x+1=0\\\Leftrightarrow\ & (\sin x-1)(\cos x-1)=0\\\Leftrightarrow\ & \left[\begin{aligned}&\sin x=1\\ &\cos x=1\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}&x=\displaystyle\frac{\pi}{2}+k2\pi\\ &x=k2\pi\end{aligned}\right., k\in \mathbb{Z}.\end{aligned}\)