$5.34 \operatorname{tg}\left(120^{\circ}+3 x\right)-\operatorname{tg}\left(140^{\circ}-x\right)=2 \sin \left(80^{\circ}+2 x\right)$.
### 5.34 We have

$$
\operatorname{tg}\left(120^{\circ}+3 x\right)+\operatorname{tg}\left(40^{\circ}+x\right)=2 \sin \left(80^{\circ}+2 x\right)
$$

Let $40^{\circ}+x=y$; then

$\operatorname{tg} 3 y+\operatorname{tg} y=2 \sin 2 y ; \cos 3 y \neq 0$, from which it follows that $\cos y \neq 0$; $\frac{\sin 4 y}{\cos y \cos 3 y}=2 \sin 2 y ; \frac{\sin 2 y \cos 2 y}{\cos y \cos 3 y}=\sin 2 y$.

## From this we obtain:

1) $\sin 2 y=0$; since $\cos y \neq 0$, then $\sin y=0$, i.e., $y_{1}=180^{\circ} \cdot k$;
2) $\cos 2 y=\cos y \cos 3 y ; 2 \cos 2 y=\cos 4 y+\cos 2 y ; \cos 2 y-$

$-\cos 4 y=0 ; \sin 3 y \sin y=0 ; \sin 3 y=0, y_{2}=60^{\circ} \cdot k$ (this solution includes $y_{1}$).

Answer: $x=60^{\circ} \cdot k-40^{\circ}, k \in Z$.