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