## 1. Solve the equation

$$
\sin ^{2} x+\sin 2 x \sin 4 x+\sin 3 x \sin 9 x+\ldots+\sin n x \sin n^{2} x=1
$$
Solution. Using the formula

$$
\sin \alpha \sin \beta=\frac{1}{2}(\cos (\alpha-\beta)-\cos (\alpha+\beta))
$$

the given equation takes the form:

$$
\frac{1}{2}[\cos 2 x-\cos 6 x+\cos 6 x-\cos 12 x+\ldots+\cos n(n-1) x-\cos n(n+1) x]=1-\sin ^{2} x
$$

i.e.

$$
\cos n(n+1) x=-1
$$

from which we get $n(n+1) x=(2 k+1) \pi$, i.e. $x=\frac{2 k+1}{n(n+1)} \pi, \quad k \in Z$.