Answer
$$x= \frac{\pi}{10}+ \frac{2k\pi}{5}$$
Work Step by Step
Given $$ \sin 3x \cos 2x + \cos3x \sin2 x=1 $$
Then by using $\sin(A+B)= \sin A \cos B+ \cos A \sin B$ , we get
\begin{align*}
\sin 3x \cos 2x + \cos3x \sin2 x&=1\\
\sin(5x)&=1\\
\end{align*}
We find all possible solutions for $x$
$$ 5x=\frac{\pi}{2}+2k\pi \ \Rightarrow x= \frac{\pi}{10}+ \frac{2k\pi}{5}$$
