$5.30 \operatorname{tg} x \operatorname{tg} 20^{\circ}+\operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ}+\operatorname{tg} 40^{\circ} \operatorname{tg} x=1$.

## Group B

Solve the equations ( $5.31-5.58$ ):
5.30 After factoring out $\operatorname{tg} x$ and further transformations, we get

$$
\begin{aligned}
& \operatorname{tg} x\left(\operatorname{tg} 20^{\circ}+\operatorname{tg} 40^{\circ}\right)=1-\operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ} ; \operatorname{tg} x=\frac{1-\operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ}}{\operatorname{tg} 20^{\circ}+\operatorname{tg} 40^{\circ}} \\
& \operatorname{tg} x=\frac{1}{\operatorname{tg}\left(20^{\circ}+40^{\circ}\right)} ; \operatorname{tg} x=\operatorname{ctg} 60^{\circ} ; \operatorname{tg} x=\operatorname{tg} 30^{\circ} ; \\
& x=30^{\circ}+180^{\circ} \cdot k
\end{aligned}
$$

Answer: $x=30^{\circ}+180^{\circ} \cdot k, k \in \boldsymbol{Z}$.