Prove that: $\sin 3x + \sin 2x - \sin x = 4\sin x\cos \dfrac{x}{2}\cos \dfrac{{3x}}{2}$

Hint: For a question like this we approach the solution by simplifying anyone the side and proving it equal to the other side, here also we will simplify the left-hand side using some of the trigonometric formulas like
$\sin A - \sin B = 2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)$
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$\sin 2A = 2\sin A\cos A$
We simplify in such a manner that it results in the equivalent value to the other side expression.

Complete step by step Answer:

Given data:$\sin 3x + \sin 2x - \sin x = 4\sin x\cos \dfrac{x}{2}\cos \dfrac{{3x}}{2}$
On simplifying the left-hand side
$ \Rightarrow \sin 3x + \sin 2x - \sin x$
On rearranging we get,
$ \Rightarrow \sin 3x - \sin x + \sin 2x$
Using the formula $\sin A - \sin B = 2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)$
$ \Rightarrow 2\sin \left( {\dfrac{{3x - x}}{2}} \right)\cos \left( {\dfrac{{3x + x}}{2}} \right) + \sin 2x$
On simplification we get,
$ \Rightarrow 2\sin \left( x \right)\cos \left( {2x} \right) + \sin 2x$
Now using the formula$\sin 2A = 2\sin A\cos A$
$ \Rightarrow 2\sin \left( x \right)\cos \left( {2x} \right) + 2\sin x\cos x$
Taking 2sinx common from both the terms,
$ \Rightarrow 2\sin x(\cos 2x + \cos x)$
Now using the formula$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$, we get,
$ \Rightarrow 2\sin x\left[ {2\cos \left( {\dfrac{{2x + x}}{2}} \right)\cos \left( {\dfrac{{2x - x}}{2}} \right)} \right]$
$ \Rightarrow 2 \times 2\cos \left( {\dfrac{{3x}}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)\sin x$
$ \Rightarrow 4\sin x\cos \dfrac{x}{2}\cos \dfrac{{3x}}{2}$, which is equal to the left-hand side in the given equation
Since the expression in the Left-hand sideis equal to the expression in the right-hand side
We have proved the given equation.

Note: An alternative method for the solution of this question can be
On simplifying the right-hand side
$ \Rightarrow 4\sin x\cos \dfrac{x}{2}\cos \dfrac{{3x}}{2}$
$ \Rightarrow 2\sin x\left[ {2\cos \left( {\dfrac{{2x - x}}{2}} \right)\cos \left( {\dfrac{{2x + x}}{2}} \right)} \right]$
Using the formula$2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) = \cos A + \cos B$
$ \Rightarrow 2\sin x\left[ {\cos 2x + \cos x} \right]$
On simplifying the brackets
$ \Rightarrow 2\sin x\cos 2x + 2\sin x\cos x$
Now using the formula $2\sin A\cos A = \sin 2A$
$ \Rightarrow 2\sin x\cos 2x + \sin 2x$
$ \Rightarrow 2\sin \left( {\dfrac{{3x - x}}{2}} \right)\cos \left( {\dfrac{{3x + x}}{2}} \right) + \sin 2x$
On simplifying using the formula $2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right) = \sin A - \sin B$
$ \Rightarrow [\sin 3x - \sin x] + \sin 2x$
$ \Rightarrow \sin 3x + \sin 2x - \sin x$, which is equal to the left-hand side in the given equation
Since the expression in the Left-hand side is equal to the expression in the right-hand side
We have proved the given equation