8.229. $2 \sin ^{4} t(\sin 2 t-3)-2 \sin ^{2} t(\sin 2 t-3)-1=0$.
## Solution.

Let's write the equation as

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

From this,

1) $\sin 2 t-1=0, \sin 2 t=1, 2 t=\frac{\pi}{2}+2 \pi k, t_{1}=\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k+1) k \in \mathbb{Z}$;
2) $\sin ^{2} 2 t-2 \sin 2 t-2=0$. Solving this equation as a quadratic in $\sin 2 t$, we get $\sin 2 t=1+\sqrt{3}>1, \varnothing ; \sin 2 t=1-\sqrt{3}$,

$$
2 t=(-1)^{n} \arcsin (1-\sqrt{3})+\pi n, t_{2}=(-1)^{n} \frac{1}{2} \arcsin (1-\sqrt{3})+\frac{\pi n}{2}, n \in \mathbb{Z}
$$

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