8.199. $\frac{\cos x}{\operatorname{ctg}^{2} \frac{x}{2}-\operatorname{tg}^{2} \frac{x}{2}}=\frac{1}{8} \cdot\left(1-\frac{2 \operatorname{ctg} x}{1+\operatorname{ctg}^{2} x}\right)$. 

Translate the above text into English, keeping the original text's line breaks and format:

8.199. $\frac{\cos x}{\cot^{2} \frac{x}{2}-\tan^{2} \frac{x}{2}}=\frac{1}{8} \cdot\left(1-\frac{2 \cot x}{1+\cot^{2} x}\right)$.
## Solution.

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

Write the equation in the form:

$\frac{\cos x}{\frac{\cos ^{2} \frac{x}{2}}{\sin ^{2} \frac{x}{2}}-\frac{\sin ^{2} \frac{x}{2}}{\cos ^{2} \frac{x}{2}}}=\frac{1}{8} \cdot\left(1-\frac{\frac{2 \cos x}{\sin x}}{1+\frac{\cos ^{2} x}{\sin ^{2} x}}\right) \Leftrightarrow$

$$
\begin{aligned}
& \Leftrightarrow \frac{\cos x \sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}}{\cos ^{4} \frac{x}{2}-\sin ^{4} \frac{x}{2}}=\frac{1}{8} \cdot\left(1-\frac{2 \cos x \sin ^{2} x}{\sin x\left(\sin ^{2} x+\cos ^{2} x\right)}\right) \Leftrightarrow \\
& \Leftrightarrow \frac{\cos x \sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}}{\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right)\left(\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}\right)}=\frac{1}{8}(1-2 \cos x \sin x) \Leftrightarrow \\
& \Leftrightarrow \frac{\cos x \sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}}{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}=\frac{1-\sin 2 x}{8} \Leftrightarrow \\
& \Leftrightarrow \frac{\cos x \sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}}{\cos x}=\frac{1-\sin 2 x}{8} \Leftrightarrow \sin ^{2} x=\frac{1-\sin 2 x}{2} \Leftrightarrow \\
& \Leftrightarrow 2 \sin ^{2} x-1+\sin 2 x=0 \Leftrightarrow \sin 2 x-\cos 2 x=0
\end{aligned}
$$

Dividing this equation by $\cos 2 x \neq 0$, we get $\operatorname{tg} 2 x=1, 2 x=\frac{\pi}{4}+\pi k$, $x=\frac{\pi}{8}+\frac{\pi k}{2}=\frac{\pi}{8}(4 k+1), k \in Z$

Answer: $x=\frac{\pi}{8}(4 k+1), k \in Z$