Đáp án:
$\left[\begin{array}{l}x =\dfrac{\pi}{6} + k\pi\\x= \dfrac{\pi}{2} + k\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\cos^2x - \sqrt3\sin x\cos x = 0$
$\Leftrightarrow 1 + \cos2x - \sqrt3\sin2x = 0$
$\Leftrightarrow \sqrt3\sin2x - \cos2x = 1$
$\Leftrightarrow \dfrac{\sqrt3}{2}\sin2x - \dfrac{1}{2}\cos2x = \dfrac{1}{2}$
$\Leftrightarrow \sin\left(2x - \dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{6}$
$\Leftrightarrow \left[\begin{array}{l}2x - \dfrac{\pi}{6} = \dfrac{\pi}{6} + k2\pi\\2x -\dfrac{\pi}{6} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x =\dfrac{\pi}{6} + k\pi\\x= \dfrac{\pi}{2} + k\pi\end{array}\right.\quad (k \in \Bbb Z)$
