8.318. $\frac{\tan t}{\cos ^{2} 5 t}-\frac{\tan 5 t}{\cos ^{2} t}=0$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\cos t \neq 0, \\ \cos 5 t \neq 0 .\end{array}\right.$

## Write the equation in the form

$\frac{\sin t}{\cos t \cos ^{2} 5 t}-\frac{\sin 5 t}{\cos 5 t \cos ^{2} t}=0 \Leftrightarrow \sin t \cos t-\sin 5 t \cos 5 t=0$.

Using the formula $\sin \alpha \cos \alpha=\frac{\sin 2 \alpha}{2}$, we get

$$
\frac{\sin 2 t}{2}-\frac{\sin 10 t}{2}=0, \sin 2 t-\sin 10 t=0 \Leftrightarrow-2 \cos 6 t \sin 4 t=0
$$

From this

1) $\cos 6 t=0,6 t=\frac{\pi}{2}+\pi k, t_{1}=\frac{\pi}{12}+\frac{\pi k}{6}=\frac{\pi}{12}(2 k+1), k \in Z$;
2) $\sin 4 t=0,4 t=\pi n, t_{2}=\frac{\pi n}{4}, n \in Z$.

Considering the domain of definition, $t_{2}=\pi n, n \in Z$.

Answer: $t_{1}=\frac{\pi}{12}(2 k+1) ; t_{2}=\pi n$, where $k$ and $n \in Z$.