8.203. $(\sin x+\cos x)^{4}+(\sin x-\cos x)^{4}=3-\sin 4 x$.

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

8.203. $(\sin x+\cos x)^{4}+(\sin x-\cos x)^{4}=3-\sin 4 x$.
Solution.

We have:

$\left(\sin ^{2} x+2 \sin x \cos x+\cos ^{2} x\right)^{2}+\left(\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x\right)^{2}=$

$=3-\sin 4 x \Leftrightarrow(1+\sin 2 x)^{2}+(1-\sin 2 x)^{2}=3-\sin 4 x \Leftrightarrow$

$\Leftrightarrow 1+2 \sin 2 x+\sin ^{2} 2 x+1-2 \sin 2 x+\sin ^{2} 2 x=3-\sin 4 x \Leftrightarrow$

$\Leftrightarrow 2 \sin ^{2} 2 x-1+\sin 4 x=0 \Leftrightarrow-\cos 4 x+\sin 4 x=0 \Leftrightarrow \operatorname{tg} 4 x=1$,

$4 x=\frac{\pi}{4}+\pi k, x=\frac{\pi}{16}+\frac{\pi k}{4}=\frac{\pi}{16}(4 k+1), k \in Z$.

Answer: $x=\frac{\pi}{16}(4 k+1), k \in Z$.