Answer
The required solution is $\frac{1}{2}\left[ \sin 2x-\sin x \right]$.
Work Step by Step
One of the product-to-sum formulas is $\cos \alpha \sin \beta =\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right) \right]$. So, in this question, according to the above-mentioned formula, the value of $\alpha $ is $\frac{3x}{2}$ and the value of $\beta $ is $\frac{x}{2}$.
Thus, the expression can be evaluated as provided below:
$\begin{align}
& \cos \frac{3x}{2}\sin \frac{x}{2}=\frac{1}{2}\left[ \sin \left( \frac{3x}{2}+\frac{x}{2} \right)-\sin \left( \frac{3x}{2}-\frac{x}{2} \right) \right] \\
& =\frac{1}{2}\left[ \sin \left( \frac{4x}{2} \right)-\sin \left( \frac{2x}{2} \right) \right] \\
& =\frac{1}{2}\left[ \sin 2x-\sin x \right]
\end{align}$
