Đáp án:
$S=\left\{-\dfrac{\pi}{6}+k2\pi;\dfrac{7\pi}{6}+k2\pi;\dfrac{\pi}{3}+k2\pi;-\dfrac{\pi}{3}+k2\pi\,\bigg{|}\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
$\sin2x+\cos x=(2\sin x+1)(3\cos x-1)$
$⇔2\sin x\cos x+\cos x=(2\sin x+1)(3\cos x-1)$
$⇔\cos x(2\sin x+1)=(2\sin x+1)(3\cos x-1)$
$⇔(2\sin x+1)(2\cos x-1)=0$
$⇔\left[ \begin{array}{l}2\sin x+1=0\\2\cos x-1=0\end{array} \right.⇔\left[ \begin{array}{l}\sin x=-\dfrac{1}{2}\\\cos x=\dfrac{1}{2}\end{array} \right.$
$⇔\left[ \begin{array}{l}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\\x=\dfrac{\pi}{3}+k2\pi\\x=-\dfrac{\pi}{3}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
Vậy $S=\left\{-\dfrac{\pi}{6}+k2\pi;\dfrac{7\pi}{6}+k2\pi;\dfrac{\pi}{3}+k2\pi;-\dfrac{\pi}{3}+k2\pi\,\bigg{|}\,k\in\mathbb Z\right\}$.
