Chia cả 2 vế cho 2 ta có
$\dfrac{\sqrt{3}}{2} \sin x- \dfrac{1}{2} \cos x = \dfrac{1}{\sqrt{2}}$
<->$\cos \dfrac{\pi}{6} \sin x - \sin(\dfrac{\pi}{6} \cos x = \sin \dfrac{\pi}{4}$
<->$ \sin(x-\dfrac{\pi}{4}) = \sin \dfrac{\pi}{4}$
<->$x - \dfrac{\pi}{4} = \dfrac{\pi}{4} + 2k\pi$ hoặc $x - \dfrac{\pi}{4} = 3\dfrac{\pi}{4} + 2k\pi$
<->$x = \dfrac{\pi}{2} + 2k\pi$ hoặc $x = \pi + 2k\pi$.
