8.284. $\cos ^{6} x+\sin ^{6} x-\cos ^{2} 2 x=\frac{1}{16}$. 

 

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8.284. $\cos ^{6} x+\sin ^{6} x-\cos ^{2} 2 x=\frac{1}{16}$.
## Solution.

We have

$$
\begin{aligned}
& \left(\cos ^{2} x+\sin ^{2} x\right)\left(\cos ^{4} x-\cos ^{2} x \sin ^{2} x+\sin ^{4} x\right)-\cos ^{2} 2 x-\frac{1}{16}=0 \Leftrightarrow \\
& \Leftrightarrow\left(\cos ^{2} x+\sin ^{2} x\right)^{2}-3 \cos ^{2} x \sin ^{2} x-\cos ^{2} 2 x-\frac{1}{16}=0 \Leftrightarrow \\
& \Leftrightarrow 16-12\left(4 \cos ^{2} x \sin ^{2} x\right)-16 \cos ^{2} 2 x-1=0 \Leftrightarrow \\
& \Leftrightarrow 16-12 \sin ^{2} 2 x-16 \cos ^{2} 2 x-1=0 \\
& 15-12\left(1-\cos ^{2} 2 x\right)-16 \cos ^{2} 2 x=0 \\
& 15-12+12 \cos ^{2} 2 x-16 \cos ^{2} 2 x=0, 4 \cos ^{2} 2 x=3, \cos ^{2} 2 x=\frac{3}{4} \\
& \cos 2 x= \pm \frac{\sqrt{3}}{2}, 2 x= \pm \frac{\pi}{6}+\pi k, x= \pm \frac{\pi}{12}+\frac{\pi k}{2}=\frac{\pi}{12}(6 k \pm 1), k \in Z \\
& \text { OmBem: } x=\frac{\pi}{12}(6 k \pm 1), k \in Z .
\end{aligned}
$$