8.249. $\frac{\cos ^{4} 2 x+\sin ^{4} 2 x}{\cos ^{4} 2 x-\sin ^{4} 2 x}-\frac{1}{2} \cos 4 x=\frac{\sqrt{3}}{2} \sin ^{-1} 4 x$.

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8.249. $\frac{\cos ^{4} 2 x+\sin ^{4} 2 x}{\cos ^{4} 2 x-\sin ^{4} 2 x}-\frac{1}{2} \cos 4 x=\frac{\sqrt{3}}{2} \sin ^{-1} 4 x$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\sin 4 x \neq 0, \\ \cos ^{4} 2 x-\sin ^{4} 2 x \neq 0 .\end{array}\right.$

From the condition we have

$$
\begin{aligned}
& \frac{\left(\cos ^{2} 2 x+\sin ^{2} 2 x\right)^{2}-2 \cos ^{2} 2 x \cos ^{2} 2 x}{\left(\cos ^{2} 2 x+\sin ^{2} 2 x\right)\left(\cos ^{2} 2 x-\sin ^{2} 2 x\right)}-\frac{\cos 4 x}{2}-\frac{\sqrt{3}}{2 \sin 4 x}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{2-4 \sin ^{2} 2 x \cos ^{2} 2 x}{\cos ^{2} 2 x-\sin ^{2} 2 x}-\cos 4 x-\frac{\sqrt{3}}{\sin 4 x}=0 \\
& \frac{2-\sin ^{2} 4 x}{\cos 4 x}-\cos 4 x-\frac{\sqrt{3}}{\sin 4 x}=0 \\
& \frac{2-\left(1-\cos ^{2} 4 x\right)-\cos ^{2} 4 x}{\cos 4 x}-\frac{\sqrt{3}}{\sin 4 x}=0, \frac{1}{\cos 4 x}-\frac{\sqrt{3}}{\sin 4 x}=0 \Leftrightarrow \\
& \Leftrightarrow \sin 4 x-\sqrt{3} \cos 4 x=0 \Leftrightarrow \operatorname{ctg} 4 x=\frac{1}{\sqrt{3}}, 4 x=\frac{\pi}{3}+\pi k \\
& x=\frac{\pi}{12}+\frac{\pi k}{4}=\frac{\pi}{12}(3 k+1), k \in Z \\
& \text { Answer: } x=\frac{\pi}{12}(3 k+1), k \in Z
\end{aligned}
$$