8.222. $\sin (3 \pi-x)+\tan(\pi+x)=\frac{\cos ^{-1} x-\cos x}{2 \sin x}$.
Solution.

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

From the condition we have:

$$
\begin{aligned}
& \sin x+\operatorname{tg} x=\frac{\frac{1}{\cos x}-\cos x}{2 \sin x} \Leftrightarrow \sin x+\frac{\sin x}{\cos x}=\frac{1-\cos ^{2} x}{2 \sin x \cos x} \Leftrightarrow \\
& \Leftrightarrow \frac{\sin x \cos x+\sin x}{\cos x}-\frac{\sin ^{2} x}{2 \sin x \cos x}=0 \Leftrightarrow \frac{\sin x(\cos x+1)}{\cos x}-\frac{\sin x}{2 \cos x}=0 \Leftrightarrow \\
& \Leftrightarrow \sin x\left(\frac{\cos x+1}{\cos x}-\frac{1}{2 \cos x}\right)=0, \sin x \neq 0 ; \frac{\cos x+1}{\cos x}-\frac{1}{2 \cos x}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{2 \cos x+1}{\cos x}=0 \Leftrightarrow 2 \cos x+1=0, \cos x=-\frac{1}{2}, \\
& x= \pm \frac{2}{3} \pi+2 \pi k=\frac{2}{3} \pi(3 k \pm 1), \quad k \in Z .
\end{aligned}
$$

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