8.355. $\operatorname{tg} x-\operatorname{tg} 2 x=\sin x$.

8.355. $\tan x-\tan 2 x=\sin x$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \cos 2 x \neq 0 .\end{array}\right.$

By the formula $\operatorname{tg} \alpha-\operatorname{tg} \beta=\frac{\sin (\alpha-\beta)}{\cos \alpha \cos \beta}$, we have

$$
\frac{-\sin x}{\cos x \cos 2 x}-\sin x=0 \Leftrightarrow-\sin x\left(\frac{1}{\cos x \cos 2 x}+1\right)=0
$$

## From this

1) $\sin x=0, x_{1}=\pi k, k \in Z ;$
2) $\frac{1}{\cos x \cos 2 x}+1=0 \Leftrightarrow 1+\cos x \cos 2 x=0,2 \cos ^{3} x-\cos x+1=0$,

$$
\begin{aligned}
& \left(2 \cos ^{3} x+2\right)-(\cos x+1)=0,2(\cos x+1)\left(\cos ^{2} x-\cos x+1\right)-(\cos x+1)=0 \\
& (\cos x+1)\left(2 \cos ^{2} x-2 \cos x+1\right)=0
\end{aligned}
$$

From this $\cos x+1=0, \cos x=-1, x_{2}=\pi+2 \pi n, n \in Z$;

$2 \cos ^{2} x-2 \cos x+1 \neq 0(D<0), \varnothing ; x_{2}$ is included in $x_{1}$.

Answer: $x=\pi k, k \in Z$.