Đáp án:
$\left[\begin{array}{l}x= k\pi\\x = \dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$2\sin^2x -\sqrt3\sin2x = 0$
$\Leftrightarrow 1 - \cos2x -\sqrt3\sin2x = 0$
$\Leftrightarrow \dfrac{\sqrt3}{2}\sin2x +\dfrac12\cos2x =\dfrac12$
$\Leftrightarrow \sin\left(2x +\dfrac{\pi}{6}\right) =\sin\dfrac{\pi}{6}$
$\Leftrightarrow\left[\begin{array}{l}2x +\dfrac{\pi}{6} =\dfrac{\pi}{6} + k2\pi\\2x +\dfrac{\pi}{6} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}x= k\pi\\x = \dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
