$8.488 \sin ^{10} x+\cos ^{10} x=\frac{29}{64}$.
Solution. Let $\sin ^{2} x=u, \cos ^{2} x=v$.
$\left\{\begin{array}{l}u^{5}+v^{5}=\frac{29}{64}, \Leftrightarrow u^{5}+v^{5}=(u+v)\left(u^{4}-u^{3} v+u^{2} v^{2}-u v^{3}+v^{4}\right)=\left(u^{4}+v^{4}\right)- \\ u+v=1\end{array}\right.$
$-u v \cdot\left(u^{2}+v^{2}\right)+u^{2} v^{2}=\left(u^{2}+v^{2}\right)^{2}-u v \cdot\left(u^{2}+v^{2}\right)-u^{2} v^{2}=\left((u+v)^{2}-2 u v\right)^{2}-$
$-u^{2} v^{2}-u v \cdot\left((u+v)^{2}-2 u v\right)=(1-2 u v)^{2}-u v \cdot(1-2 u v)-u^{2} v^{2}=5 u^{2} v^{2}-5 u v+1$.

Thus, $5 u^{2} v^{2}-5 u v+1=\frac{29}{64} \Leftrightarrow u^{2} v^{2}-u v+\frac{7}{64}=0 \Leftrightarrow$

$\Leftrightarrow\left(\sin ^{2} x \cdot \cos ^{2} x\right)^{2}-\sin ^{2} x \cdot \cos ^{2} x+\frac{7}{64}=0 \Leftrightarrow$

$\Leftrightarrow\left(4 \sin ^{2} x \cdot \cos ^{2} x\right)^{2}-4\left(4 \sin ^{2} x \cdot \cos ^{2} x\right)+\frac{7}{4}=0 \Leftrightarrow$

$\Leftrightarrow \sin ^{4} 2 x-4 \sin ^{2} 2 x+\frac{7}{4}=0 \Leftrightarrow\left[\begin{array}{l}\sin ^{2} 2 x=\frac{1}{2} \\ \sin ^{2} 2 x=\frac{7}{2}\end{array} \Leftrightarrow \frac{1-\cos 4 x}{2}=\frac{1}{2} \Leftrightarrow\right.$

$\Leftrightarrow \cos 4 x=0 \Leftrightarrow x=\frac{\pi}{8}+\frac{\pi k}{4}$.

Answer: $x=\frac{\pi}{8}(2 k+1)$.

$8.489(\sin x+\sqrt{3} \cos x)^{2}=5+\cos \left(\frac{\pi}{3}+4 x\right)$.

Solution.

$$
\begin{aligned}
& 4\left(\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x\right)^{2}=5+\cos \left(\frac{\pi}{3}+4 x\right) \Leftrightarrow 4 \cos ^{2}\left(x-\frac{\pi}{6}\right)=5+\cos \left(\frac{\pi}{3}+4 x\right) \Leftrightarrow \\
& \Leftrightarrow 2+2 \cos \left(2 x-\frac{\pi}{3}\right)=1+\cos \left(\frac{\pi}{3}+4 x\right)+4 \Leftrightarrow \\
& \Leftrightarrow 2 \cos \left(2 x-\frac{\pi}{3}\right)=2+2 \cos ^{2}\left(\frac{\pi}{6}+2 x\right) \Leftrightarrow \cos \left(2 x-\frac{\pi}{3}\right)=1+\cos ^{2}\left(\frac{\pi}{2}-\left(\frac{\pi}{3}-2 x\right)\right) \Leftrightarrow \\
& \Leftrightarrow \cos \left(2 x-\frac{\pi}{3}\right)=1+\sin ^{2}\left(2 x-\frac{\pi}{3}\right) \Leftrightarrow \cos ^{2}\left(2 x-\frac{\pi}{3}\right)+\cos \left(2 x-\frac{\pi}{3}\right)-2=0 \Leftrightarrow \\
& \Leftrightarrow\left[\begin{array}{l}
\cos \left(2 x-\frac{\pi}{3}\right)=-2, \\
\cos \left(2 x-\frac{\pi}{3}\right)=1
\end{array} \Leftrightarrow 2 x=\frac{\pi}{3}+2 \pi k \Leftrightarrow x=\frac{\pi}{6}+\pi k\right.
\end{aligned}
$$

Answer: $x=\frac{\pi}{6}(6 k+1)$.