Đáp án: $x\in \{\pm \dfrac{\pi}{3}+k2\pi,k\pi,\dfrac{-pi}{2}+k2\pi\}$
Giải thích các bước giải:
$2\cos^2x-3\cos x+1=\sin 2x-sin x$
$\rightarrow 2\cos^2x-2\cos x-\cos x+1=2\sin x.\cos x-sin x$
$\rightarrow (\cos x-1)(2\cos x-1)=\sin x.(2\cos x-1)$
$\rightarrow (\cos x-1-\sin x).(2\cos x-1)=0$
$+)2\cos x-1=0$
$\rightarrow \cos x=\dfrac{1}{2}$
$\rightarrow x=\pm \dfrac{\pi}{3}+k2\pi$
$+)\cos x-1-\sin x=0$
$\rightarrow \cos x=\sin x+1$
Mà $\cos^2x+\sin^2x=1$
$\rightarrow (\sin x+1)^2+\sin^2x=1$
$\rightarrow 2\sin ^2x+2\sin x=0$
$\rightarrow \sin x(\sin x+1)=0$
$+)\sin x=0\rightarrow x=k\pi$
$+)\sin x=-1\rightarrow x=\dfrac{-pi}{2}+k2\pi$
