8.177. $\operatorname{tg}^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2} \operatorname{tg} \frac{x}{2}+\cos ^{2} \frac{x}{2} \operatorname{ctg} \frac{x}{2}+\operatorname{ctg}^{2} \frac{x}{2}+\sin x=4$.
## Solution.

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

Rewrite the equation as

$$
1+\operatorname{tg}^{2} \frac{x}{2}+1+\operatorname{ctg}^{2} \frac{x}{2}+\frac{1-\cos x}{2} \cdot \frac{1-\cos x}{\sin x}+\frac{1+\cos x}{2} \cdot \frac{1+\cos x}{\sin x}+\sin x=6 \Leftrightarrow
$$

$$
\begin{aligned}
& \Leftrightarrow \frac{1}{\cos ^{2} \frac{x}{2}}+\frac{1}{\sin ^{2} \frac{x}{2}}+\frac{1-2 \cos x+\cos ^{2} x+1+2 \cos x+\cos ^{2} x}{2 \sin x}+ \\
& +\sin x=6 \Leftrightarrow \frac{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}}{\sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}}+\frac{2+2-2 \sin ^{2} x}{2 \sin x}+\sin x-6=0 \Leftrightarrow \\
& \Leftrightarrow \frac{4}{\sin ^{2} x}+\frac{2 \cdot}{\sin x}-\sin x+\sin x-6=0 \Leftrightarrow \frac{2}{\sin ^{2} x}+\frac{1}{\sin x}-3=0 \Leftrightarrow \\
& \Leftrightarrow 3 \cdot \sin ^{2} x-\sin x-2=0(\sin x \neq 0) .
\end{aligned}
$$

Solving this equation as a quadratic in $\sin x$, we get $(\sin x)_{1}=-\frac{2}{3}, x_{1}=(-1)^{k+1} \arcsin \frac{2}{3}+\pi k, k \in Z ;$ $(\sin x)_{2}=1, x_{2}=\frac{\pi}{2}+2 \pi n=\frac{\pi}{2}(4 n+1), n \in Z ;$

Answer: $x_{1}=(-1)^{k+1} \arcsin \frac{2}{3}+\pi k ; x_{2}=\frac{\pi}{2}(4 n+1)$, where $k$ and $n \in Z$.