Giải thích các bước giải:

Ta có:

$\cos2x+\sqrt3\sin2x=\sqrt3\cos x-\sin x$

$\to \dfrac12\cos2x+\dfrac{\sqrt3}2\sin2x=\dfrac{\sqrt3}2\cos x-\dfrac12\sin x$

$\to \cos(\dfrac{\pi}3)\cos2x+\sin(\dfrac{\pi}3)\sin2x=\cos(\dfrac{\pi}6)\cos x-\sin(\dfrac{\pi}6)\sin x$

$\to \cos(2x-\dfrac{\pi}3)=\cos(x+\dfrac{\pi}6)$

$\to 2x-\dfrac{\pi}3=x+\dfrac{\pi}6+k2\pi\to x=\dfrac12\pi+k2\pi, k\in Z$

Hoặc $2x-\dfrac{\pi}3=-(x+\dfrac{\pi}6)+k2\pi\to x=\dfrac{1}{18}\pi+\dfrac23k\pi$