8.207. $2 \cos ^{6} 2 t-\cos ^{4} 2 t+1.5 \sin ^{2} 4 t-3 \sin ^{2} 2 t=0$.
Solution.

Since $\cos ^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{2}$ and $\sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2}$, we have

$2\left(\frac{1}{2}(1+\cos 4 t)\right)^{3}-\left(\frac{1}{2}(2+\cos 4 t)\right)^{2}+1.5\left(1-\cos ^{2} 4 t\right)-1.5(1-\cos 4 t)=0 \Leftrightarrow$ $\Leftrightarrow \frac{(1+\cos 4 t)^{3}}{4}-\frac{(1+\cos 4 t)^{2}}{4}+\frac{3\left(1-\cos ^{2} 4 t\right)}{2}-\frac{3(1-\cos 4 t)}{2}=0 \Rightarrow$ $\Rightarrow \cos ^{3} 4 t-4 \cos ^{2} 4 t+7 \cos 4 t=0 \Leftrightarrow \cos 4 t\left(\cos ^{2} 4 t-4 \cos 4 t+7\right)=0$.

From this, 1) $\cos 4 t=0, 4 t=\frac{\pi}{2}+\pi k, t=\frac{\pi}{8}+\frac{\pi k}{4}=\frac{\pi}{8}(2 k+1), k \in Z$;

$$
\text { 2) } \cos ^{2} 4 t-4 \cos 4 t+7 \neq 0(D<0), \varnothing
$$

Answer: $t=\frac{\pi}{8}(2 k+1), k \in Z$.