8.268. $\operatorname{tg} \frac{3 x}{2}-\operatorname{tg} \frac{x}{2}=2 \sin x$.

8.268. $\tan \frac{3 x}{2}-\tan \frac{x}{2}=2 \sin x$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\cos \frac{3 x}{2} \neq 0, \\ \cos \frac{x}{2} \neq 0 .\end{array}\right.$

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

$\frac{\sin x}{\cos \frac{3 x}{2} \cos \frac{x}{2}}-2 \sin x=0 \Leftrightarrow \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{\cos \frac{3 x}{2} \cos \frac{x}{2}}-4 \sin \frac{x}{2} \cos \frac{x}{2}=0 \Leftrightarrow$

$\Leftrightarrow 2 \sin \frac{x}{2} \cdot\left(\frac{1}{\cos \frac{3 x}{2}}-2 \cos \frac{x}{2}\right)=0 \Leftrightarrow \sin \frac{x}{2}\left(1-2 \cos \frac{3 x}{2} \cos \frac{x}{2}\right)=0 \Leftrightarrow$

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

From this,

1) $\sin \frac{x}{2}=0, \frac{x}{2}=\pi k, x_{1}=2 \pi k, k \in Z$;
2) $2 \cos ^{2} x+\cos x-2=0$. Solving the equation as a quadratic in terms of $\cos x$, we get $\cos x=\frac{-1-\sqrt{17}}{4}<-1, \varnothing$; or $\cos x=\frac{-1+\sqrt{17}}{4}$, $x_{2}= \pm \arccos \frac{-1+\sqrt{17}}{4}+2 \pi n, n \in Z$.

Answer: $x_{1}=2 \pi k, x_{2}= \pm \arccos \frac{-1+\sqrt{17}}{4}+2 \pi n$, where $k$ and $n \in Z$.