8.055. $\cos ^{4} 2 x+6 \cos ^{2} 2 x=\frac{25}{16}$.
## Solution.

We have $16 \cos ^{4} 2 x+96 \cos ^{2} 2 x-25=0$. Solving this equation as a biquadratic equation in terms of $\cos 2 x$, we get $\cos 2 x= \pm \frac{1}{2}, 2 x= \pm \frac{\pi}{3}+\pi k$, $x= \pm \frac{\pi}{6}+\frac{\pi k}{2}$, where $k \in Z$.

Answer: $\quad x= \pm \frac{\pi}{6}+\frac{\pi k}{2}, k \in Z$.