2cos ²x-3cosx+1=sin2x-sinx. Ai giải hộ vs

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 $\cos^2x+\sin^2x=1$

$\rightarrow (\sin x+1)^2+\sin^2x=1$

$\rightarrow 2\sin ^2x+2\sin x=0$

$\rightarrow \sin x(\sin x+1)=0$

$+)\sin x=0\rightarrow x=k\pi$

$+)\sin x=-1\rightarrow x=\dfrac{-pi}{2}+k2\pi$