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