8.367. $\frac{\tan 2 t}{\cos ^{2} t}-\frac{\tan t}{\cos ^{2} 2 t}=0$.
Solution.

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

Write the equation in the form

$\frac{\sin 2 t}{\cos 2 t \cos ^{2} t}-\frac{\sin t}{\cos t \cos ^{2} 2 t}=0, \frac{\sin 2 t \cos 2 t-\sin t \cos t}{\cos ^{2} t \cos ^{2} 2 t}=0 \Leftrightarrow$ $\Leftrightarrow \frac{2 \sin t \cos t \cos 2 t-\sin t \cos t}{\cos ^{2} t \cos ^{2} 2 t}=0 \Leftrightarrow \frac{\sin t \cos t(2 \cos 2 t-1)}{\cos ^{2} t \cos ^{2} 2 t}=0$, $\frac{\sin t(2 \cos 2 t-1)}{\cos t \cos ^{2} 2 t}=0 \Leftrightarrow \sin t(2 \cos 2 t-1)=0$.

Then

1) $\sin t=0, t_{1}=\pi k, k \in Z$;
2) $2 \cos 2 t-1=0, \cos 2 t=\frac{1}{2}, 2 t= \pm \frac{\pi}{3}+2 \pi n, t_{2}= \pm \frac{\pi}{6}+\pi n, n \in Z$.

Answer: $t_{1}=\pi k, t_{2}= \pm \frac{\pi}{6}+\pi n$, where $k$ and $n \in Z$.