Đáp án:
$\left[\begin{array}{l}x =\dfrac{\pi}{2} +k2\pi\\x= \dfrac{\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\cos2x + \sin x = \sqrt3(\cos x - \sin2x)\\ \Leftrightarrow \cos2x + \sqrt3\sin2x = \sqrt3\cos x - \sin x\\ \Leftrightarrow \dfrac{1}{2}\cos2x + \dfrac{\sqrt3}{2}\sin2x = \dfrac{\sqrt3}{2}\cos x - \dfrac{1}{2}\sin x\\ \Leftrightarrow \cos\left(2x - \dfrac{\pi}{3}\right) = \cos\left(x + \dfrac{\pi}{6}\right)\\ \Leftrightarrow \left[\begin{array}{l}2x - \dfrac{\pi}{3} = x + \dfrac{\pi}{6} +k2\pi\\2x - \dfrac{\pi}{3} = - x - \dfrac{\pi}{6} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =\dfrac{\pi}{2} +k2\pi\\x= \dfrac{\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
