8.231. $2 \sin ^{4} x+1.25 \sin ^{2} 2 x-\cos ^{4} x=\cos 2 x$.
## Solution.

We have $8\left(\sin ^{2} x\right)^{2}+5 \sin ^{2} 2 x-4\left(\cos ^{2} x\right)^{2}-4 \cos 2 x=0$. Using the formulas

$$
\begin{aligned}
& \sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2} \text { and } \cos ^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{2}, \text { we find } \\
& 8\left(\frac{1}{2}(1-\cos 2 x)\right)^{2}+5\left(1-\cos ^{2} 2 x\right)-4\left(\frac{1}{2}(1+\cos 2 x)\right)^{2}-4 \cos 2 x=0 \Leftrightarrow \\
& \quad \Leftrightarrow 2(1-\cos 2 x)^{2}+5(1-\cos 2 x)^{2}-\left(1+\cos ^{2} 2 x\right)-4 \cos 2 x=0 \Leftrightarrow \\
& \Leftrightarrow 2-4 \cos 2 x+2 \cos ^{2} 2 x+5-5 \cos ^{2} 2 x-1-2 \cos 2 x-\cos ^{2} 2 x- \\
& -4 \cos 2 x=0 \Leftrightarrow 2 \cos ^{2} 2 x+5 \cos 2 x-3=0
\end{aligned}
$$

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

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