Chia 2 vế cho 2 ta có
$\dfrac{\sqrt{3}}{2} \sin x - \dfrac{1}{2} \cos x = \sin(2x)$
$<-> \sin x \cos(\dfrac{\pi}{6}) - \cos x \sin(\dfrac{\pi}{6}) = \sin(2x)$
$<-> \sin(x - \dfrac{\pi}{6}) = \sin(2x)$
Vậy $2x = x - \dfrac{\pi}{6} + 2k\pi$ hoặc $2x = \pi - x +\dfrac{\pi}{6} + 2k\pi$.
Do đó $x = -\dfrac{\pi}{6} + 2k\pi$ hoặc $x = \dfrac{7\pi}{18} + \dfrac{2k\pi}{3}$.