8.321. $\sin ^{2}\left(t+45^{\circ}\right)-\sin ^{2}\left(t-30^{\circ}\right)-\sin 15^{\circ} \cos \left(2 t+15^{\circ}\right)=0.5 \sin 6 t$.
## Solution.

From the condition, we have

$$
\begin{aligned}
& 1-\cos \left(2 t+90^{\circ}\right)-1+\cos \left(2 t-60^{\circ}\right)-2 \sin 15^{\circ} \cos \left(2 t+15^{\circ}\right)-\sin 6 t=0 \Leftrightarrow \\
& \Leftrightarrow \sin 2 t+\cos \left(2 t-60^{\circ}\right)+\sin 2 t-\sin \left(2 t+30^{\circ}\right)-\sin 6 t=0 \Leftrightarrow \\
& \Leftrightarrow 2 \sin 2 t+\cos 2 t \cos 60^{\circ}+\sin 2 t \sin 60^{\circ}-\sin 2 t \cos 30^{\circ}- \\
& -\cos 2 t \sin 30^{\circ}-\sin 6 t=0,
\end{aligned}
$$

$2 \sin 2 t+\frac{1}{2} \cos 2 t+\frac{\sqrt{3}}{2} \sin 2 t-\frac{\sqrt{3}}{2} \sin 2 t-\frac{1}{2} \cos 2 t-\sin 6 t=0$,

$2 \sin 2 t-\sin 6 t=0$.

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

From this,

1) $\sin 2 t=0, 2 t=180^{\circ} k, t_{1}=90^{\circ} k, k \in \mathbb{Z}$;
2) $4 \sin ^{2} 2 t-1=0, \sin 2 t= \pm \frac{1}{2}, 2 t= \pm 30^{\circ}+180^{\circ} l, t_{2}= \pm 15^{\circ}+90^{\circ} l, l \in \mathbb{Z}$.

Answer: $t_{1}=90^{\circ} k, t_{2}= \pm 15^{\circ}+90^{\circ} l$, where $k$ and $l \in \mathbb{Z}$.