8.072. $(\cos 6 x-1) \operatorname{ctg} 3 x=\sin 3 x$.

8.072. $(\cos 6 x-1) \cot 3 x=\sin 3 x$.
## Solution.

Domain of definition: $\sin 3 x \neq 0$.

We have

$\frac{(\cos 2(3 x)-1) \cos 3 x}{\sin 3 x}-\sin 3 x=0, \quad\left(2 \cos ^{2} 3 x-2\right) \cos 3 x-\sin ^{2} 3 x=0$,

$-2\left(1-\cos ^{2} 3 x\right) \cos 3 x-\sin ^{2} 3 x=0,2 \sin ^{2} 3 x \cos 3 x+\sin ^{2} 3 x=0$,

$\sin ^{2} 3 x(2 \cos 3 x+1)=0$.

Since $\sin 3 x \neq 0$, then $2 \cos 3 x+1=0, \cos 3 x=-\frac{1}{2}, 3 x= \pm \frac{2}{3} \pi+2 \pi k$, $x= \pm \frac{2}{9} \pi+\frac{2}{3} \pi k, k \in Z$

Answer: $\quad x= \pm \frac{2}{9} \pi+\frac{2}{3} \pi k, k \in Z$