8.061. $4 \operatorname{tg}^{2} 3 x-\cos ^{-2} 3 x=2$.

Translate to English:

8.061. $4 \tan^{2} 3 x-\cos ^{-2} 3 x=2$.
## Solution.

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

## We have

$$
\frac{4 \sin ^{2} 3 x}{\cos ^{2} 3 x}-\frac{1}{\cos ^{2} 3 x}-2=0 \Leftrightarrow \frac{4\left(1-\cos ^{2} 3 x\right)}{\cos ^{2} 3 x}-\frac{1}{\cos ^{2} 3 x}-2=0
$$

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

From this, $\cos 3 x= \pm \frac{\sqrt{2}}{2}, 3 x= \pm \frac{\pi}{4}+\pi k, x= \pm \frac{\pi}{12}+\frac{\pi k}{3}, k \in Z$.

Answer: $\quad x= \pm \frac{\pi}{12}+\frac{\pi k}{3}, k \in Z$.