3. Solve the trigonometric equation

$$
\sin \left(\frac{\pi}{10}+\frac{3 x}{2}\right)=2 \sin \left(\frac{3 \pi}{10}-\frac{x}{2}\right)
$$
Solution. Writing the given equation in the form

$$
\sin \left(\frac{\pi}{10}+\frac{3 x}{2}\right)-\sin \left(\frac{3 \pi}{10}-\frac{x}{2}\right)=\sin \left(\frac{3 \pi}{10}-\frac{x}{2}\right)
$$

and using the formula for the difference of sines, we obtain

$$
2 \cos \left(\frac{x}{2}+\frac{\pi}{5}\right) \sin \left(x-\frac{\pi}{10}\right)=\sin \left(\frac{3 \pi}{10}-\frac{x}{2}\right)
$$

Using the trigonometric identity $\cos \alpha=\sin \left(\frac{\pi}{2}-\alpha\right)$, the equation is transformed into

$$
2 \sin \left(\frac{3 \pi}{10}-\frac{x}{2}\right) \sin \left(x-\frac{\pi}{10}\right)=\sin \left(\frac{3 \pi}{10}-\frac{x}{2}\right)
$$

i.e.,

$$
\sin \left(\frac{3 \pi}{10}-\frac{x}{2}\right)\left[2 \sin \left(x-\frac{\pi}{10}\right)-1\right]=0
$$

From the last equation, it follows that

$$
\sin \left(\frac{3 \pi}{10}-\frac{x}{2}\right)=0 \text { or } \sin \left(x-\frac{\pi}{10}\right)=\frac{1}{2} \text {. }
$$

From the first equation, we get

$$
\frac{3 \pi}{10}-\frac{X}{2}=k \pi, k \in \mathbb{Z}
$$

and from the second

$$
x-\frac{\pi}{10}=\frac{\pi}{6}+2 k \pi, x-\frac{\pi}{10}=\frac{5 \pi}{6}+2 k \pi, k \in \mathbb{Z}
$$

Thus, the solutions to the given equation are:

$$
x=\frac{3+10 k}{5} \pi, x=\frac{4+30 k}{15} \pi, x=\frac{14+30 k}{15} \pi, k \in \mathbb{Z} .
$$