3. Solve the equation

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

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

equation (1) is transformed into

$$
2-\cos 2 x-\cos 4 x+2 \cos ^{2} 3 x=2
$$

and then, considering that

$$
\cos \alpha+\cos \beta=2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}
$$

from (2) we get

$$
-2 \cos 3 x \cos x+2 \cos ^{2} 3 x=0
$$

i.e.

$$
(-\cos x+\cos 3 x) \cos 3 x=0
$$

Since $\cos 3 x-\cos x=-2 \sin 2 x \sin x$, equation (3) takes the form

$$
\sin x \sin 2 x \cos 3 x=0
$$

so

$$
\sin x=0, \sin 2 x=0, \cos 3 x=0
$$

The solutions to the first, second, and third equations in (4) are respectively:

$$
x=k \pi, x=k \frac{\pi}{2}, x=(2 k+1) \frac{\pi}{2},(k \in \mathbb{Z})
$$

these are all the solutions to the given equation (1).