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

Let's write the equation as $\operatorname{tg} 3\left(x+40^{\circ}\right)-\operatorname{tg}\left(180^{\circ}-\left(x+40^{\circ}\right)\right)=$

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

Using the formula $\operatorname{tg} \alpha+\operatorname{tg} \beta=\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}$ and letting $x+40^{\circ}=y$, we can write

$$
\begin{aligned}
& \frac{\sin 4 y}{\cos 3 y \cos y}-2 \sin 2 y=0 \Leftrightarrow \frac{\sin 4 y-2 \sin 2 y \cos 3 y \cos y}{\cos 3 y \cos y}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{2 \sin 2 y \cos 2 y-2 \sin 2 y \cos 3 y \cos y}{\cos 3 y \cos y}=0 \Leftrightarrow \frac{\sin 2 y(2 \cos 2 y-2 \cos 3 y \cos y)}{\cos 3 y \cos y}= \\
& =0 \Leftrightarrow \frac{2 \sin y \cos y(2 \cos 2 y-2 \cos 3 y \cos y)}{\cos 3 y \cos y}=0, \quad(\cos y \neq 0) \Rightarrow \\
& \Rightarrow \frac{\sin y(2 \cos 2 y-2 \cos 3 y \cos y)}{\cos 3 y}=0
\end{aligned}
$$

Since $\cos \alpha \cos \beta=\frac{1}{2}(\cos (\alpha-\beta)+\cos (\alpha+\beta))$, we have

$$
\frac{\sin y(2 \cos 2 y-\cos 2 y-\cos 4 y)}{\cos 3 y}=0 \Leftrightarrow \frac{\sin y(\cos 2 y-\cos 4 y)}{\cos 3 y}=0 .
$$

Using the formula $\cos \alpha-\cos \beta=-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$, we get

$\frac{-2 \sin y \sin 3 y \sin y}{\cos 3 y}=0, \frac{\sin ^{2} y \sin 3 y}{\cos 3 y}=0 \Rightarrow \sin y=0$,

or $\sin 3 y=0(\cos 3 y \neq 0)$.

The solutions to the equation $\sin y=0$ are included in the solutions to the equation

$$
\sin 3 y=0\left(\sin 3 y=3 \sin y-4 \sin ^{3} y=\sin y\left(3-4 \sin ^{2} y\right)\right)
$$

from which $3 y=180^{\circ} k, y=60^{\circ} k, k \in Z, x+40^{\circ}=60^{\circ} k, x=-40^{\circ}+60^{\circ} k, k \in Z$.

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