$8.435 \sin ^{10} 3 x+\cos ^{10} 3 x=4 \frac{\sin ^{6} 3 x+\cos ^{6} 3 x}{4 \cos ^{2} 6 x+\sin ^{2} 6 x}$.
Solution. Transform the right side of the original equation:

$4 \frac{\sin ^{6} 3 x+\cos ^{6} 3 x}{4 \cos ^{2} 6 x+\sin ^{2} 6 x}=\frac{4\left(\frac{(1-\cos 6 x)^{3}}{2^{3}}+\frac{(1+\cos 6 x)^{3}}{2^{3}}\right)}{1+3 \cos ^{2} 6 x}=\frac{1}{2} \cdot \frac{2+6 \cos ^{2} 6 x}{3 \cos ^{2} 6 x+1}=1$.

From here $\sin ^{10} 3 x+\cos ^{10} 3 x=1 \Leftrightarrow \sin ^{10} 3 x+\cos ^{10} 3 x=\sin ^{2} 3 x+\cos ^{2} 3 x \Leftrightarrow$
$\Leftrightarrow \sin ^{2} 3 x \cdot\left(1-\sin ^{8} 3 x\right)+\cos ^{2} 3 x \cdot\left(1-\cos ^{8} 3 x\right)=0 \Leftrightarrow\left\{\begin{array}{l}\sin ^{2} 3 x \cdot\left(1-\sin ^{8} 3 x\right)=0, \\ \cos ^{2} 3 x \cdot\left(1-\cos ^{8} 3 x\right)=0\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}{\left[\begin{array}{l}\sin 3 x=0, \\ \sin 3 x= \pm 1,\end{array}\right.} \\ {\left[\begin{array}{l}\cos 3 x=0, \\ \cos 3 x= \pm 1\end{array}\right.}\end{array} \Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}\sin 3 x=0, \\ \cos 3 x= \pm 1,\end{array}\right. \\ \left\{\begin{array}{l}\cos 3 x=0, \\ \sin 3 x= \pm 1\end{array}\right.\end{array} \Leftrightarrow 3 x=\frac{\pi k}{2}\right.\right.$.

Answer: $\quad=\frac{\pi k}{6}$.

## $8.436 \operatorname{ctg}^{4} 2 x=\cos ^{2} 4 x+1$.

Solution. $\operatorname{ctg}^{4} 2 x-1=\cos ^{2} 4 x \Leftrightarrow\left(\operatorname{ctg}^{2} 2 x-1\right)\left(\operatorname{ctg}^{2} 2 x+1\right)=\cos ^{2} 4 x \Leftrightarrow$ $\Leftrightarrow \frac{\cos 4 x}{\sin ^{4} 2 x}=\cos ^{2} 4 x \Leftrightarrow\left\{\begin{array}{l}\cos 4 x \cdot\left(1-\cos 4 x \cdot \sin ^{4} 2 x\right)=0, \\ \sin 2 x \neq 0 .\end{array}\right.$

1) $\left\{\begin{array}{l}\cos 4 x=0, \\ \sin 2 x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}2 x=\frac{\pi}{4}+\frac{\pi k}{2}, \\ 2 x \neq \pi k\end{array} \Leftrightarrow x=\frac{\pi}{8}+\frac{\pi k}{4}\right.\right.$.
2) $1-\cos 4 x \cdot \sin ^{4} 2 x=0 \Leftrightarrow\left\{\begin{array}{c}\cos 4 x=1 \\ \sin ^{4} 2 x=1\end{array} \Leftrightarrow\left\{\begin{array}{l}2 \sin ^{2} 2 x=0 \\ \sin ^{2} 2 x=1\end{array} \Leftrightarrow \varnothing\right.\right.$.

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

$8.437\left(2+\frac{1}{\cos ^{2} x}\right)\left(4-2 \cos ^{4} x\right)=1+5 \sin 3 y$.

Solution. $2+\frac{1}{\cos ^{2} x} \geq 3,4-2 \cos ^{4} x \geq 2 \Rightarrow$ the left side of the equation is not less than 6, while the right side does not exceed 6. Therefore, the original equation is equivalent to:

$$
\left\{\begin{array} { l } 
{ 1 + 5 \operatorname { s i n } 3 y = 6 } \\
{ 4 - 2 \operatorname { c o s } ^ { 4 } x = 2 , } \\
{ 2 + \frac { 1 } { \operatorname { c o s } ^ { 2 } x } = 3 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \operatorname { s i n } 3 y = 1 , } \\
{ \operatorname { c o s } ^ { 2 } x = 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y=\frac{\pi}{6}+\frac{2 \pi n}{3} \\
x=\pi k
\end{array}\right.\right.\right.
$$

Answer: $x=\pi k, y=\frac{\pi}{6}(4 n+1)$.