Answer
$$x= \frac{3\pi}{10}+ \frac{2k\pi}{5}$$
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 $x$
$$ 5x=\frac{3\pi}{2}+2k\pi \ \Rightarrow x= \frac{3\pi}{10}+ \frac{2k\pi}{5}$$