Answer
See the explanation below.
Work Step by Step
We use the trigonometric identity: $2{{\cos }^{2}}x-1=\cos 2x$.
Now, the left side can be written as:
$\begin{align}
& \sin x(4{{\cos }^{2}}x-1)=\sin x(2{{\cos }^{2}}x+2{{\cos }^{2}}x-1) \\
& =\sin x(2{{\cos }^{2}}x+\cos 2x) \\
& =2\sin x{{\cos }^{2}}x+\sin x\cos 2x \\
& =2\sin x\cos x\cos x+\sin x\cos 2x
\end{align}$
We use the trigonometric identities,
$2\sin x\cos x=\sin 2x$ And $\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta $
Simplified further,
$\begin{align}
& 2\sin x\cos x\cos x+\sin x\cos 2x=\sin 2x\cos x+\sin x\cos 2x \\
& =\sin (x+2x) \\
& =\sin 3x
\end{align}$
Thus, the left side of the equation is equal to $\sin 3x$.
Hence, the left side is equal to $\sin 3x$.
