8.315. $4\left(\sin t \cos ^{5} t+\cos t \sin ^{5} t\right)+\sin ^{3} 2 t=1$.
## Solution.

We have

$4 \sin t \cos t\left(\cos ^{4} t+\sin ^{4} t\right)+\sin ^{3} 2 t-1=0$,

$2 \sin 2 t\left(\left(\cos ^{2} t+\sin ^{2} t\right)^{2}-2 \cos ^{2} t \sin ^{2} t\right)+\sin ^{3} 2 t-1=0$,

$2 \sin 2 t-\sin 2 t\left(4 \cos ^{2} t \sin ^{2} t\right)+\sin ^{3} 2 t-1=0$

$2 \sin 2 t-\sin ^{3} 2 t+\sin ^{3} 2 t-1=0, \sin 2 t=\frac{1}{2}, 2 t=(-1)^{k} \frac{\pi}{6}+\pi k$, $t=(-1)^{k} \frac{\pi}{12}+\frac{\pi k}{2}, k \in Z$

Answer: $t=(-1)^{k} \frac{\pi}{12}+\frac{\pi k}{2}, k \in Z$