Ta có:
$\cos 2x-\sqrt{3}\sin 2x-\sqrt{3}\sin x-\cos x+4=0$
$\Leftrightarrow \dfrac{1}{2}\cos 2x-\dfrac{\sqrt{3}}{2}\sin 2x-\dfrac{\sqrt{3}}{2}\sin x-\dfrac{1}{2}\cos x+2=0$
$\Leftrightarrow \cos(2x+\dfrac{\pi}{3})-\sin(x+\dfrac{\pi}{6})+2=0$
$\Leftrightarrow 1-2\sin^2(x+\dfrac{\pi}{6})-\sin(x+\dfrac{\pi}{6})+2=0$
$\Leftrightarrow 2\sin^2(x+\dfrac{\pi}{6})+\sin(x+\dfrac{\pi}{6})-3=0$
$\Leftrightarrow \sin(x+\dfrac{\pi}{6})=1$
$\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi,k\in\mathbb{Z}$
$\Leftrightarrow x=\dfrac{\pi}{3}+k2\pi,k\in\mathbb{Z}$