8.290. $\sin ^{4} 3 t+\sin ^{4}\left(\frac{\pi}{4}+3 t\right)=\frac{1}{4}$
Solution.

Write the equation in the form

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

$\Leftrightarrow\left(\frac{1}{2} \cdot(1-\cos 6 t)\right)^{2}+\left(\frac{1}{2} \cdot\left(1-\cos \left(\frac{\pi}{2}+6 t\right)\right)\right)^{2}-\frac{1}{4}=0 \Leftrightarrow$

$\Leftrightarrow(1-\cos 6 t)^{2}+(1+\sin 6 t)^{2}-1=0$,

$1-2 \cos 6 t+\cos ^{2} 6 t+1+2 \sin 6 t+\sin ^{2} 6 t-1=0,1-\cos 6 t+\sin 6 t=0 \Leftrightarrow$

$\Leftrightarrow \cos ^{2} 3 t+\sin ^{2} 3 t-\cos ^{2} 3 t+\sin ^{2} 3 t+2 \sin 3 t \cos 3 t=0$,

$2 \sin ^{2} 3 t+2 \sin 3 t \cos 3 t=0,2 \sin 3 t \cdot(\sin 3 t+\cos 3 t)=0$.

From this 1) $\sin 3 t=0,2) \sin 3 t+\cos 3 t=0$.

1) $3 t=\pi k, t_{1}=\frac{\pi k}{3}, k \in Z$
2) $\operatorname{tg} 3 t=-1,3 t=-\frac{\pi}{4}+\pi n, t_{2}=-\frac{\pi}{12}+\frac{\pi n}{3}=\frac{\pi}{12}(4 n-1), n \in Z$.

Answer: $t_{1}=\frac{\pi k}{3} ; t_{2}=\frac{\pi}{12}(4 n-1)$, where $k$ and $n \in Z$.