8.221. $\operatorname{tg} 3 t+\operatorname{tg} t=2 \sin 4 t$.

8.221. $\tan 3t + \tan t = 2 \sin 4t$.
Solution.

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

Since $\operatorname{tg} \alpha+\operatorname{tg} \beta=\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}$, we can rewrite the equation as $\frac{\sin (3 t+t)}{\cos 3 t \cos t}-2 \sin 4 t=0 \Leftrightarrow \frac{\sin 4 t-\sin 4 t(2 \cos 3 t \cos t)}{\cos 3 t \cos t}=0 \Leftrightarrow$

$\Leftrightarrow \sin 4 t(1-2 \cos 3 t \cos t)=0$.

Applying the formula $\cos \alpha \cos \beta=\frac{1}{2}(\cos (\alpha-\beta)+\cos (\alpha+\beta))$, we have $\sin 4 t(1-\cos 2 t-\cos 4 t)=0 \Leftrightarrow \sin 4 t\left(1-\cos 2 t-2 \cos ^{2} 2 t+1\right)=0 \Leftrightarrow$ $\Leftrightarrow \sin 4 t\left(2 \cos ^{2} 2 t+\cos 2 t-2\right)=0$.

From this,

1) $\sin 4 t=0, 4 t=\pi k, t_{1}=\frac{\pi k}{4}, k \in Z$;
2) $2 \cos ^{2} 2 t+\cos 2 t-2=0$. Solving this equation as a quadratic equation in terms of $\cos 2 t$, we find $\cos 2 t=\frac{-1-\sqrt{17}}{4}<-1, \varnothing$; or $\cos 2 t=\frac{-1+\sqrt{17}}{4}$, $2 t= \pm \arccos \frac{\sqrt{17}-1}{4}+2 \pi n, t_{2}= \pm \frac{1}{2} \arccos \frac{\sqrt{17}-1}{4}+\pi n, n \in Z$. Considering the domain of definition, $t_{1}=\frac{\pi k}{4}, k \neq 4 l+2$, where $k$ and $l \in Z$.

Answer: $t_{1}=\frac{\pi k}{4}, k \neq 4 l+2 ; t_{2}= \pm \frac{1}{2} \arccos \frac{\sqrt{17}-1}{4}+\pi n$, where $k, l$ and $n \in Z$