8.314. $\frac{2 \sin x-\sin 2 x}{2 \sin x+\sin 2 x}+\operatorname{ctg}^{2} \frac{x}{2}=\frac{10}{3}$.
## Solution.

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

Rewrite the equation as

$$
\begin{aligned}
& \frac{2 \sin x-2 \sin x \cos x}{2 \sin x+2 \sin x \cos x}+\frac{1+\cos x}{1-\cos x}-\frac{10}{3}=0 \\
& \frac{2 \sin x(1-\cos x)}{2 \sin x(1+\cos x)}+\frac{1+\cos x}{1-\cos x}-\frac{10}{3}=0, \frac{1-\cos x}{1+\cos x}+\frac{1+\cos x}{1-\cos x}-\frac{10}{3}=0 \\
& 3 \cdot\left(\frac{1-\cos x}{1+\cos x}\right)^{2}-10 \cdot\left(\frac{1-\cos x}{1+\cos x}\right)+3=0
\end{aligned}
$$

Solving this equation as a quadratic in $\frac{1-\cos x}{1+\cos x}$, we get

1) $\left(\frac{1-\cos x}{1+\cos x}\right)=\frac{1}{3}, \cos x=\frac{1}{2}, x_{1}= \pm \frac{\pi}{3}+2 \pi l, l \in Z$;
2) $\left(\frac{1-\cos x}{1+\cos x}\right)=3, \cos x=-\frac{1}{2}, x_{2}= \pm \frac{2}{3} \pi+2 \pi n, n \in Z$;

$$
x= \pm \frac{\pi}{3}+\pi k=\frac{\pi}{3}(3 k \pm 1), k \in Z
$$

Answer: $x=\frac{\pi}{3}(3 k \pm 1), k \in Z$.