8.100. $1-\cos 6 x=\tan 3 x$.
## Solution.

Domain of definition: $\cos 3 x \neq 0, 3 x \neq \frac{\pi}{2}+\pi n, x \neq \frac{\pi}{6}(2 n+1), n \in Z$.

## We have

$1-\cos 6 x-\frac{\sin 6 x}{1+\cos 6 x}=0, \quad(1-\cos 6 x)(1+\cos 6 x)-\sin 6 x=0$,

$1-\cos ^{2} 6 x-\sin 6 x=0, \sin ^{2} 6 x-\sin 6 x=0, \sin 6 x(\sin 6 x-1)=0$.

Then:

1) $\sin 6 x=0, 6 x=\pi n, x_{1}=\frac{\pi n}{6}$, but considering the domain of definition $x_{1}=\frac{\pi m}{3}$, $m \neq 2 n+1, m \in Z$
2) $\sin 6 x-1=0, \sin 6 x=1, 6 x=\frac{\pi}{2}+2 \pi k, x_{2}=\frac{\pi}{12}+\frac{\pi k}{3}=\frac{\pi}{12}(4 k+1)$, $k \in Z$.

Answer: $\quad x_{1}=\frac{\pi m}{3} ; x_{2}=\frac{\pi}{12}(4 k+1), m, k \in Z$.