8.247. $\frac{\cos ^{3} 3 t}{\tan t}+\frac{\cos ^{2} t}{\tan 3 t}=0$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\cos t \neq 0, \\ \cos 3 t \neq 0, \\ \tan t \neq 0, \\ \tan 3 t \neq 0 .\end{array}\right.$

Rewrite the equation as

$$
\begin{aligned}
& \frac{\cos ^{2} 3 t \cos t}{\sin t}+\frac{\cos ^{2} t \cos 3 t}{\sin 3 t}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\cos ^{2} 3 t \cos t \sin 3 t+\cos ^{2} t \cos 3 t \sin t}{\sin t \sin 3 t}=0 \Leftrightarrow
\end{aligned}
$$

$$
\Leftrightarrow \frac{\cos 3 t \cos t(\sin 3 t \cos 3 t+\sin t \cos t)}{\sin t \sin 3 t}=0 \Leftrightarrow
$$

$\Leftrightarrow \cos 3 t \cos t(\sin 3 t \cos 3 t+\sin t \cos t)=0$.

Since $\cos 3 t \neq 0, \cos t \neq 0$, then

$\sin 3 t \cos 3 t+\sin t \cos t=0, \frac{\sin 6 t}{2}+\frac{\sin 2 t}{2}=0$,

$\sin 6 t+\sin 2 t=0 \Leftrightarrow 2 \sin 4 t \cos 2 t=0$.

Then

$\sin 4 t=0,4 t=\pi l, t_{1}=\frac{\pi l}{4}, l \in Z ; \cos 2 t=0,2 t=\frac{\pi}{2}+\pi m$,

$t_{2}=\frac{\pi}{4}+\frac{\pi m}{2}=\frac{\pi}{4}(2 m+1), m \in Z ; t_{1}$ does not fit the domain of definition when $l=2 k$, and when $l=2 k+1$ we get the series $t_{2}$.

Answer: $t=\frac{\pi}{4}(2 m+1), m \in Z$.