4. Solve the equation $\frac{2 \operatorname{tg}^{4} 6 x+4 \sin 4 x \sin 8 x-\cos 8 x-\cos 16 x+2}{\sqrt{\cos x-\sqrt{3} \sin x}}=0$.
# Solution:

$$
\frac{2 \operatorname{tg}^{4} 6 x+4 \sin 4 x \sin 8 x-\cos 8 x-\cos 16 x+2}{\sqrt{\cos x-\sqrt{3} \sin x}}=0
$$

Given the condition $\cos x-\sqrt{3} \sin x>0$, we find the roots of the equation $2 \operatorname{tg}^{4} 6 x+4 \sin 4 x \sin 8 x-\cos 8 x-\cos 16 x+2=0 \Leftrightarrow$
$2 \operatorname{tg}^{4} 6 x+4 \sin 4 x \sin 8 x-1+2 \sin ^{2} 4 x-1+2 \sin ^{2} 8 x+2=0 \Leftrightarrow 2 \operatorname{tg}^{4} 6 x+2(\sin 4 x+\sin 8 x)^{2}=0 \Leftrightarrow$ $\left\{\begin{array}{c}\operatorname{tg} 6 x=0, \\ \sin 4 x+\sin 8 x=0,\end{array} \Leftrightarrow\left\{\begin{array}{c}x=\frac{\pi n}{6}, \\ {\left[\begin{array}{c}\cos 2 x=0, \\ \sin 6 x=0,\end{array}\right.}\end{array} \Leftrightarrow\left\{\begin{array}{c}x=\frac{\pi n}{6}, n \in Z, \\ {\left[\begin{array}{c}x=\frac{\pi}{4}+\frac{\pi k}{2}, k \in Z, \\ x=\frac{\pi}{6}, l \in Z,\end{array} \Leftrightarrow x=\frac{\pi l}{6}, l \in Z .\right.}\end{array}\right.\right.\right.$

Considering the condition $\cos x-\sqrt{3} \sin x>0$, we finally have $x=2 \pi n, x=-\frac{\pi}{6}+2 \pi n$, $x=-\frac{\pi}{3}+2 \pi n, x=-\frac{\pi}{2}+2 \pi n, x=-\frac{2 \pi}{3}+2 \pi n, n \in Z$.

Answer: $x=2 \pi n, x=-\frac{\pi}{6}+2 \pi n, x=-\frac{\pi}{3}+2 \pi n, x=-\frac{\pi}{2}+2 \pi n, x=-\frac{2 \pi}{3}+2 \pi n, n \in Z$.