8.059. $2 \operatorname{tg}^{4} 3 x-3 \operatorname{tg}^{2} 3 x+1=0$.

8.059. $2 \tan^{4} 3 x-3 \tan^{2} 3 x+1=0$.
## Solution.

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

Solving this equation as a biquadratic equation in terms of $\operatorname{tg} 3 x$, we get: 1) $\left.\operatorname{tg} 3 x= \pm \frac{\sqrt{2}}{2}, x_{1,2}=\frac{ \pm \operatorname{arctg} \frac{\sqrt{2}}{2}}{3}+\frac{\pi k}{3}, k \in Z ; 2\right) \operatorname{tg} 3 x= \pm 1$, $x_{3,4}= \pm \frac{\pi}{12}+\frac{\pi n}{3}, n \in Z$

Answer: $\quad x_{1,2}= \pm \frac{1}{3} \operatorname{arctg} \frac{\sqrt{2}}{2}+\frac{\pi k}{3} ; x_{3,4}= \pm \frac{\pi}{12}+\frac{\pi n}{3}, k, n \in Z$