8.007. $\operatorname{tg} 3 t-\operatorname{tg} t-4 \sin t=0$.

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8.007. $\tan 3 t-\tan t-4 \sin t=0$.
Solution.

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

Using the formula $\operatorname{tg} \alpha-\operatorname{tg} \beta=\frac{\sin (\alpha-\beta)}{\cos \alpha \cos \beta}$, rewrite the equation as $\frac{\sin 2 t}{\cos 3 t \cos t}-4 \sin t=0, \frac{2 \sin t \cos t}{\cos 3 t \cos t}-4 \sin t=0, \Leftrightarrow 2 \sin t \cdot\left(\frac{1-2 \cos 3 t}{\cos 3 t}\right)=0$. From this, $\sin t=0, t_{1}=\pi k, k \in Z$, or $1-2 \cos 3 t=0, \cos 3 t=\frac{1}{2}, 3 t= \pm \frac{\pi}{3}+2 \pi n$, $t_{2}= \pm \frac{\pi}{9}+\frac{2 \pi n}{3}, n \in Z$

Answer: $t_{1}=\pi k ; t_{2}= \pm \frac{\pi}{9}+\frac{2 \pi n}{3}, \quad k, n \in Z$