8.213. $2 \sin ^{5} 2 t-\sin ^{3} 2 t-6 \sin ^{2} 2 t+3=0$.

8.213. $2 \sin ^{5} 2 t-\sin ^{3} 2 t-6 \sin ^{2} 2 t+3=0$.
## Solution.

Let's write the equation as:

$$
\begin{aligned}
& \sin ^{3} 2 t\left(2 \sin ^{2} 2 t-1\right)-3\left(2 \sin ^{2} 2 t-1\right)=0 \Leftrightarrow \\
& \Leftrightarrow\left(2 \sin ^{2} 2 t-1\right)\left(\sin ^{3} 2 t-3\right)=0
\end{aligned}
$$

## From this

1) $2 \sin ^{2} 2 t-1=0, \sin 2 t= \pm \frac{\sqrt{2}}{2}, 2 t=\frac{\pi}{4}+\frac{\pi k}{2}$,

$$
t_{1}=\frac{\pi}{8}+\frac{\pi k}{4}=\frac{\pi}{8}(2 k+1), \quad k \in Z_{i}
$$

2) $\sin ^{3} 2 t-3=0, \sin 2 t=\sqrt[3]{3}>1, \varnothing$.

Answer: $t=\frac{\pi}{8}(2 k+1), k \in Z$.