8.354. $\operatorname{tg}\left(\frac{3 \pi}{2}-x\right)+\frac{\cos \left(\frac{7 \pi}{2}+x\right)}{1+\cos x}=2$.
## Solution.

$\left\{\begin{array}{l}\sin x \neq 0 \\ \cos x \neq-1\end{array}\right.$

## We have

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

From this

1) $1+\cos x=0, \cos x=-1, x_{1}=\pi+2 \pi k, k \in Z$;
2) $1-2 \sin x=0, \sin x=\frac{1}{2}, x_{2}=(-1)^{\prime} \frac{\pi}{6}+\pi l, l \in Z ; x_{1}$ does not fit the domain.

Answer: $x=(-1)^{\prime} \frac{\pi}{6}+\pi l, l \in Z$.