4. Solve the equation $\frac{\sin 2 x-\cos 2 x+\sqrt{3} \cos x+\sqrt{3} \sin x+1}{\sqrt{\sqrt{3} \cos x-\sin x}}=0 . \quad(8$ points)
# Solution:

$\frac{\sin 2 x-\cos 2 x+\sqrt{3} \cos x+\sqrt{3} \sin x+1}{\sqrt{\sqrt{3} \cos x-\sin x}}=0 \Leftrightarrow \frac{\sin 2 x+2 \sin ^{2} x+\sqrt{3} \cos x+\sqrt{3} \sin x}{\sqrt{\sqrt{3} \cos x-\sin x}}=0$.

Given the condition $\sqrt{3} \cos x-\sin x>0$, we find the roots of the equation

$\sin 2 x+2 \sin ^{2} x+\sqrt{3} \cos x+\sqrt{3} \sin x=0 \Leftrightarrow 2 \sin x \cos x+2 \sin ^{2} x+\sqrt{3} \cos x+\sqrt{3} \sin x=0 \Leftrightarrow$

$2 \sin x(\cos x+\sin x)+\sqrt{3}(\cos x+\sin x)=0 \quad \Leftrightarrow \quad(2 \sin x+\sqrt{3})(\cos x+\sin x)=0 \quad \Leftrightarrow$

$2 \sin x+\sqrt{3}=0, \quad \sin x+\cos x=0 \Leftrightarrow \sin x=-\frac{\sqrt{3}}{2}, \quad \operatorname{tg} x=-1 \Leftrightarrow$

$x=-\frac{\pi}{3}+2 \pi n, \quad x=-\frac{2 \pi}{3}+2 \pi n, \quad x=-\frac{\pi}{4}+\pi n n \in Z$. Considering the condition $\sqrt{3} \cos x-\sin x>0$, we finally have $x=-\frac{\pi}{3}+2 \pi n, x=-\frac{\pi}{4}+2 \pi n, n \in Z$.

Answer: $x=-\frac{\pi}{3}+2 \pi n, x=-\frac{\pi}{4}+2 \pi n, n \in Z$.