$8.442 \sin ^{8} 2 x+\cos ^{8} 2 x=\frac{41}{128}$. 

 

The translation is as follows:

$8.442 \sin ^{8} 2 x+\cos ^{8} 2 x=\frac{41}{128}$.
## Solution.

$\left(\sin ^{4} 2 x-\cos ^{4} 2 x\right)^{2}+2 \cos ^{4} 2 x \cdot \sin ^{4} 2 x=\frac{41}{128} \Leftrightarrow 8 \cos ^{2} 4 x+\sin ^{4} 4 x=\frac{41}{16} \Leftrightarrow$ $\Leftrightarrow \sin ^{4} 4 x-8 \sin ^{2} 4 x+\frac{87}{16}=0 \Leftrightarrow \sin ^{2} 4 x=4 \pm \frac{13}{4} \Leftrightarrow \sin ^{2} 4 x=\frac{3}{4}$

$\Leftrightarrow x= \pm \frac{\pi}{12}+\frac{\pi k}{4}$.

Answer: $x=\frac{\pi}{12}(3 k \pm 1)$.

## $8.4432(1-\sin x-\cos x)+\tan x+\cot x=0$.

## Solution.

$(\tan x+2+\cot x)-2(\sin x+\cos x)=0 \Leftrightarrow \frac{(\sin x+\cos x)^{2}}{\sin x \cdot \cos x}-2(\sin x+\cos x)=0 \Leftrightarrow$

1) $\sin x=-\cos x \Leftrightarrow \tan x=-1, x=-\frac{\pi}{4}+\pi k$.
2) $\sin x+\cos x-2 \sin x \cdot \cos x=0 \Leftrightarrow(\sin x+\cos x)+1-(1+2 \sin x \cdot \cos x)=0 \Leftrightarrow$ $\Leftrightarrow(\sin x+\cos x)+1-(\sin x+\cos x)^{2}=0 \Leftrightarrow 2 \cos ^{2}\left(x-\frac{\pi}{4}\right)-\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)-1=0 \Leftrightarrow$ $\Leftrightarrow \cos \left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2} \pm \sqrt{10}}{4} \Rightarrow \cos \left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}-\sqrt{10}}{4}$.

Answer: $x_{1}=\frac{\pi}{4} \pm \arccos \frac{\sqrt{2}-\sqrt{10}}{4}+2 \pi k, x_{2}=-\frac{\pi}{4}+\pi k$.