8.261. $(2 \cos 2 t+5) \cos ^{4} t-(2 \cos 2 t+5) \sin ^{4} t=3$.
## Solution.

From the condition we have:

$$
\begin{aligned}
& (2 \cos 2 t+5)\left(\cos ^{4} t-\sin ^{4} t\right)-3=0 \Leftrightarrow \\
& \Leftrightarrow(2 \cos 2 t+5)\left(\cos ^{2} t+\sin ^{2} t\right)\left(\cos ^{2} t-\sin ^{2} t\right)-3=0 \Leftrightarrow \\
& \Leftrightarrow(2 \cos 2 t+5) \cos 2 t-3=0 \Leftrightarrow 2 \cos ^{2} 2 t+5 \cos 2 t-3=0
\end{aligned}
$$

Solving this equation as a quadratic equation in terms of $\cos 2 t$, we find $\cos 2 t=-3, \varnothing ; \cos 2 t=\frac{1}{2}, 2 t= \pm \frac{\pi}{3}+2 \pi k, t= \pm \frac{\pi}{6}+\pi k=\frac{\pi}{6}(6 k \pm 1), k \in Z$.

Answer: $t=\frac{\pi}{6}(6 k \pm 1), k \in Z$.