8.359. $37 \operatorname{tg} 3x = 11 \operatorname{tg} x$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\cos 3 x \neq 0, \\ \cos x \neq 0 .\end{array}\right.$

Using the formula $\operatorname{tg} 3 \alpha=\frac{3 \operatorname{tg} \alpha-\operatorname{tg}^{3} \alpha}{1-3 \operatorname{tg}^{2} \alpha}$, rewrite the equation as

$$
37 \cdot \frac{3 \operatorname{tg} x-\operatorname{tg}^{3} x}{1-3 \operatorname{tg}^{2} x}-11 \operatorname{tg} x=0 \Leftrightarrow \operatorname{tg} x \cdot\left(\frac{111-37 \operatorname{tg}^{2} x-11+33 \operatorname{tg}^{2} x}{1-3 \operatorname{tg}^{2} x}\right)=0 \Leftrightarrow
$$

$$
\Leftrightarrow \operatorname{tg} x \cdot\left(\frac{100-4 \operatorname{tg}^{2} x}{1-3 \operatorname{tg}^{2} x}\right)=0 \Leftrightarrow\left\{\begin{array} { l } 
{ [ \begin{array} { l } 
{ \operatorname { tg } x = 0 } \\
{ \operatorname { tg } ^ { 2 } x = 2 5 }
\end{array} } \\
{ \operatorname { tg } ^ { 2 } x \neq \frac { 1 } { 3 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
{\left[\begin{array}{l}
\operatorname{tg} x=0 \\
\operatorname{tg} x= \pm 5
\end{array}\right.} \\
\operatorname{tg} x \neq \pm \frac{1}{\sqrt{3}}
\end{array}\right.\right.
$$

The last solution does not satisfy the original equation. Therefore, $x_{1}=\pi k, k \in Z ; x_{2}=\operatorname{arctg}( \pm 5)+\pi n= \pm \operatorname{arctg} 5+\pi n, n \in Z$.

Answer: $x_{1}=\pi k, k \in Z ; x_{2}= \pm \operatorname{arctg} 5+\pi n, n \in Z$.