8.323. $\frac{5 \sin x-5 \tan x}{\sin x+\tan x}+4(1-\cos x)=0$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0, \\ \cos x \neq-1 .\end{array}\right.$

From the condition we have

$$
\begin{aligned}
& \frac{5 \sin x-\frac{5 \sin x}{\cos x}}{\sin x+\frac{\sin x}{\cos x}}+4(1-\cos x)=0 \Leftrightarrow \\
& \Leftrightarrow \frac{5 \sin x(\cos x-1)}{\cos x} \cdot \frac{\cos x}{\sin x(\cos x+1)}+4(1-\cos x)=0 \\
& (\cos x-1)\left(\frac{5}{\cos x+1}-4\right)=0
\end{aligned}
$$

## From here

1) $\cos x-1=0, \cos x=1, x_{1}=2 \pi k, k \in Z$;
2) $\frac{5}{\cos x+1}-4=0, \cos x=\frac{1}{4}, x_{2}= \pm \arccos \frac{1}{4}+2 \pi n, n \in Z ; x_{1}$ does not satisfy the domain of definition.

Answer: $x= \pm \arccos \frac{1}{4}+2 \pi n, n \in Z$.