4. Solve the equation $\frac{4 \operatorname{tg}^{4} 8 x+4 \sin 2 x \sin 6 x-\cos 4 x-\cos 12 x+2}{\sqrt{\cos x-\sin x}}=0$.
# Solution:

$$
\frac{4 \operatorname{tg}^{4} 8 x+4 \sin 2 x \sin 6 x-\cos 4 x-\cos 12 x+2}{\sqrt{\cos x-\sin x}}=0
$$

Given the condition $\cos x-\sin x>0$, we find the roots of the equation $4 \operatorname{tg}^{4} 8 x+4 \sin 2 x \sin 6 x-\cos 4 x-\cos 12 x+2=0 \Leftrightarrow$
$4 \operatorname{tg}^{4} 8 x+4 \sin 2 x \sin 6 x-1+2 \sin ^{2} 2 x-1+2 \sin ^{2} 6 x+2=0 \Leftrightarrow 4 \operatorname{tg}^{4} 8 x+2(\sin 2 x+\sin 6 x)^{2}=0 \Leftrightarrow$ $\left\{\begin{array}{c}\operatorname{tg} 8 x=0, \\ \sin 2 x+\sin 6 x=0,\end{array} \Leftrightarrow\left\{\begin{array}{c}x=\frac{\pi n}{8}, \\ {\left[\begin{array}{c}\cos 2 x=0, \\ \sin 4 x=0,\end{array}\right.}\end{array} \Leftrightarrow\left\{\begin{array}{c}x=\frac{\pi n}{8}, n \in Z, \\ {\left[\begin{array}{c}x=\frac{\pi}{4}+\frac{\pi k}{2}, k \in Z, \\ x=\frac{\pi l}{4}, l \in Z,\end{array} \Leftrightarrow x=\frac{\pi l}{4}, l \in Z .\right.}\end{array}\right.\right.\right.$

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

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