Đáp án:
$\left[\begin{array}{l}x =\dfrac{\pi}{12} + k\pi\\x = -\dfrac{\pi}{4} + k\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$2\sqrt3\cos2x - 2\sin2x = 2$
$\Leftrightarrow \dfrac{\sqrt3}{2}\cos2x - \dfrac{1}{2}\sin2x = \dfrac{1}{2}$
$\Leftrightarrow \cos\left(2x +\dfrac{\pi}{6}\right) = \cos\dfrac{\pi}{3}$
$\Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{6} = \dfrac{\pi}{3} + k2\pi\\2x + \dfrac{\pi}{6} = -\dfrac{\pi}{3} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x =\dfrac{\pi}{12} + k\pi\\x = -\dfrac{\pi}{4} + k\pi\end{array}\right.\quad (k \in \Bbb Z)$
