8.030. $2+\tan x \cot \frac{x}{2}+\cot x \tan \frac{x}{2}=0$.
Solution.

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

Using the half-angle formulas $\operatorname{ctg} \frac{\alpha}{2}=\frac{1+\cos \alpha}{\sin \alpha}, \operatorname{tg} \frac{\alpha}{2}=\frac{\sin \alpha}{1+\cos \alpha}$, therefore

$$
\begin{aligned}
& 2+\frac{\sin x}{\cos x} \cdot \frac{1+\cos x}{\sin x}+\frac{\cos x}{\sin x} \cdot \frac{\sin x}{1+\cos x}=0 \Rightarrow 2+\frac{1+\cos x}{\cos x}+\frac{\cos x}{1+\cos x}=0 \\
& 4 \cos ^{2} x+4 \cos x+1=0,(2 \cos x+1)^{2}=0
\end{aligned}
$$

from which $\cos x=-\frac{1}{2}, \quad x= \pm \frac{2}{3} \pi+2 \pi k=\frac{2 \pi}{3}(3 k \pm 1), \quad k \in Z$.

Answer: $\quad x=\frac{2 \pi}{3}(3 k \pm 1), k \in Z$.