8.272. $\sin ^{3} 2 t+\cos ^{3} 2 t+\frac{1}{2} \sin 4 t=1$.
## Solution.

Let's write the equation in the form

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

$\Leftrightarrow(\sin 2 t+\cos 2 t)(1-\sin 2 t \cos 2 t)-(1-\sin 2 t \cos 2 t)=0 \Leftrightarrow$

$\Leftrightarrow(1-\sin 2 t \cos 2 t)(\sin 2 t+\cos 2 t-1)=0$.

From this,

I) $1-\sin 2 t \cos 2 t=0, \frac{1}{2} \sin 4 t=1, \sin 4 t=2, \varnothing$

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

$\sin t \cos t-2 \sin ^{2} t=0,2 \sin t(\cos t-\sin t)=0 \Rightarrow$

1) $\sin t=0, t_{1}=\pi k, k \in Z$
2) $\cos t-\sin t=0, \operatorname{tg} t=1, t_{2}=\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n+1), n \in Z$.

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