8.242. $\frac{1}{\tan 5 x+\tan 2 x}-\frac{1}{\cot 5 x+\cot 2 x}=\tan 3 x$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\cos 3 x \neq 0, \\ \operatorname{tg} 5 x+\operatorname{tg} 2 x \neq 0, \\ \operatorname{ctg} 5 x+\operatorname{ctg} 2 x \neq 0, \\ \cos 5 x \neq 0, \\ \cos 2 x \neq 0, \\ \sin 5 x \neq 0, \\ \sin 2 x \neq 0 .\end{array}\right.$

Since $\operatorname{tg} \alpha+\operatorname{tg} \beta=\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}, \operatorname{ctg} \alpha+\operatorname{ctg} \beta=\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}$, we have $\frac{\cos 5 x \cos 2 x}{\sin 7 x}-\frac{\sin 5 x \sin 2 x}{\sin 7 x}-\frac{\sin 3 x}{\cos 3 x}=0 \Leftrightarrow$

$\Leftrightarrow \frac{\cos 5 x \cos 2 x-\sin 5 x \sin 2 x}{\sin 7 x}-\frac{\sin 3 x}{\cos 3 x}=0 \Leftrightarrow \frac{\cos 7 x}{\sin 7 x}-\frac{\sin 3 x}{\cos 3 x}=0$, $\frac{\cos 7 x \cos 3 x-\sin 3 x \sin 7 x}{\sin 7 x \cos 3 x}=0 \Leftrightarrow \frac{\cos 10 x}{\sin 7 x \cos 3 x}=0 \Rightarrow \cos 10 x=0$, $\sin 7 x \cos 3 x \neq 0,10 x=\frac{\pi}{2}+\pi k, x=\frac{\pi}{20}+\frac{\pi k}{10}=\frac{\pi}{20}(2 k+1), k \neq 5 l+2$, since otherwise $\cos 2 x=0$.

Answer: $x=\frac{\pi}{20}(2 k+1), k \neq 5 l+2, k$ and $l \in Z$.