8.236. $\frac{1}{\sin ^{3} \frac{x}{2} \cos ^{3} \frac{x}{2}}-6 \cos ^{-1} x=\tan^{3} \frac{x}{2}+\cot^{3} \frac{x}{2}$.
## Solution.

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

Since $\sin \alpha=2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}, \operatorname{tg} \frac{\alpha}{2}=\frac{1-\cos \alpha}{\sin \alpha}, \operatorname{ctg} \frac{\alpha}{2}=\frac{1+\cos \alpha}{\sin \alpha}$, the given equation can be written as

$$
\begin{aligned}
& \frac{8}{\sin ^{3} x}-\frac{6}{\cos x}=\frac{(1-\cos x)^{3}}{\sin ^{3} x}+\frac{(1+\cos x)^{3}}{\sin ^{3} x} . \\
& 8 \cos x-6 \sin ^{3} x=\left(1-3 \cos x+3 \cos ^{2} x-\cos ^{3} x+1+3 \cos x+3 \cos ^{2} x+\right. \\
& \left.+\cos ^{3} x\right) \cos x \Leftrightarrow 8 \cos x-6 \sin ^{3} x-2 \cos x-6 \cos ^{3} x=0 \Leftrightarrow \\
& \Leftrightarrow \sin ^{3} x+\cos ^{3} x-\cos x=0 \Leftrightarrow \sin ^{3} x-\cos x \sin ^{2} x=0 \Leftrightarrow \\
& \Leftrightarrow \sin ^{2} x(\sin x-\cos x)=0 .
\end{aligned}
$$

Since $\sin x \neq 0$, then $\sin x-\cos x=0, \sin x=\cos x, \operatorname{tg} x=1, x=\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k+1), k \in Z$. Answer: $x=\frac{\pi}{4}(4 k+1), k \in Z$.