8.133. $\sin ^{2} 3 x+\sin ^{2} 4 x=\sin ^{2} 5 x+\sin ^{2} 6 x$.

8.133. $\sin ^{2} 3 x+\sin ^{2} 4 x=\sin ^{2} 5 x+\sin ^{2} 6 x$. 

The equation remains the same in English as it is a mathematical expression.
## Solution.

Using the formulas for reducing the power

$$
\begin{aligned}
& \frac{1}{2}(1-\cos 6 x)+\frac{1}{2}(1-\cos 8 x)=\frac{1}{2}(1-\cos 10 x)+\frac{1}{2}(1-\cos 12 x) \\
& (\cos 6 x+\cos 8 x)-(\cos 10 x+\cos 12 x)=0 \Leftrightarrow 2 \cos \frac{6 x+8 x}{2} \cos \frac{6 x-8 x}{2}- \\
& -2 \cos \frac{10 x+12 x}{2} \cos \frac{10 x-12 x}{2}=0 \Leftrightarrow 2 \cos 7 x \cos x-2 \cos 11 x \cos x=0, \\
& 2 \cos x(\cos 7 x-\cos 11 x)=0
\end{aligned}
$$

From this:

1) $\cos x=0, \quad x_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), \quad k \in Z$;
2) $\cos 7 x-\cos 11 x=0 \Leftrightarrow 2 \sin \frac{7 x+11 x}{2} \sin \frac{11 x-7 x}{2}=0$,

$\sin 9 x \sin 2 x=0$.

From this:
a) $\sin 9 x=0,9 x=\pi n, x_{2}=\frac{\pi n}{9}, n \in Z$;

b) $\sin 2 x=0,2 x=\pi l, x_{3}=\frac{\pi l}{2}, l \in Z ; x_{1}$ is included in $x_{3}$.

Answer: $\quad x_{1}=\frac{\pi l}{2} ; x_{2}=\frac{\pi n}{9}, l, n \in Z$.