If $\sin 3\alpha =4\sin \alpha \sin (x+\alpha )\sin (x-\alpha )$ then $x=$.
A. \[n\pi \pm \dfrac{\pi }{6}\]
B. \[n\pi \pm \dfrac{\pi }{3}\]
C. \[n\pi \pm \dfrac{\pi }{4}\]
D. \[n\pi \pm \dfrac{\pi }{2}\]

Hint: To find the value of $x$ we will use the formula of $\sin 3A$ and identity of $\sin (A+B)\sin (A-B)$in the given equation. We will then simplify the equation and derive an equation where we will apply the theorem according to which for all the real numbers $x$ and $y$, $\sin x=\sin y$ implies that $x=n\pi \pm y$ where $n$ is an integer.

Formula Used: $\sin 3A=3\sin A-4{{\sin }^{3}}A$
$\sin (A+B)\sin (A-B)={{\sin }^{2}}A-{{\sin }^{2}}B$

Complete step by step solution: Complete step-by-step solution:
We are given a trigonometric equation $\sin 3\alpha =4\sin \alpha \sin (x+\alpha )\sin (x-\alpha )$ and we have to determine the value of $x$.
We will use the formula of $\sin 3A$ in the equation.
$\begin{align}
  & \sin 3\alpha =4\sin \alpha \sin (x+\alpha )\sin (x-\alpha ) \\
 & 3\sin \alpha -4{{\sin }^{3}}\alpha =4\sin \alpha \left[ \sin (x+\alpha )\sin (x-\alpha ) \right]
\end{align}$
Now as we know that $\sin (A+B)\sin (A-B)={{\sin }^{2}}A-{{\sin }^{2}}B$, we will use this in the equation.
$\begin{align}
  & 3\sin \alpha -4{{\sin }^{3}}\alpha =4\sin \alpha \left[ {{\sin }^{2}}x-{{\sin }^{2}}\alpha \right] \\
 & \sin \alpha (3-4{{\sin }^{2}}\alpha )=4\sin \alpha \left[ {{\sin }^{2}}x-{{\sin }^{2}}\alpha \right] \\
 & 3-4{{\sin }^{2}}\alpha =4{{\sin }^{2}}x-4{{\sin }^{2}}\alpha \\
 & 3=4{{\sin }^{2}}x \\
 & {{\sin }^{2}}x=\dfrac{3}{4} \\
 & \sin x=\pm \dfrac{\sqrt{3}}{2}
\end{align}$
We know that $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$. SO,
$\sin x=\sin \left( \pm \dfrac{\pi }{3} \right)$
Applying the theorem here we will get,
$x=n\pi \pm \dfrac{\pi }{3}$, here $n\in Z$.
The value of $x$ for the trigonometric equation $\sin 3\alpha =4\sin \alpha \sin (x+\alpha )\sin (x-\alpha )$ is $x=n\pi \pm \dfrac{\pi }{3}$

Option ‘B’ is correct

Note: We have used the formula or identity $\sin (A+B)\sin (A-B)={{\sin }^{2}}A-{{\sin }^{2}}B$ directly in the question. This identity is derived from the formula of $\sin (a+b)=\sin a\cos b+\cos a\sin b$ and $\sin (a-b)=\sin a\cos b-\cos a\sin b$. So instead of using $\sin (A+B)\sin (A-B)={{\sin }^{2}}A-{{\sin }^{2}}B$ directly, we can also derive the equation by using these formulas.
 Here we will show.
$\sin (x+\alpha )\sin (x-\alpha )=\left[ \sin x\cos \alpha +\cos x\sin \alpha \right]\left[ \sin x\cos \alpha -\cos x\sin \alpha \right]$
Using formula of $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$,
\[\sin (x+\alpha )\sin (x-\alpha )={{\sin }^{2}}x{{\cos }^{2}}\alpha -{{\cos }^{2}}x{{\sin }^{2}}\alpha \]
We will now use the formula ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ in the above equation.
\[\begin{align}
  & \sin (x+\alpha )\sin (x-\alpha )={{\sin }^{2}}x\left( 1-{{\sin }^{2}}\alpha \right)-\left( 1-{{\sin }^{2}}x \right){{\sin }^{2}}\alpha \\
 & ={{\sin }^{2}}x-{{\sin }^{2}}x{{\sin }^{2}}\alpha -{{\sin }^{2}}\alpha +{{\sin }^{2}}x{{\sin }^{2}}\alpha \\
 & ={{\sin }^{2}}x-{{\sin }^{2}}\alpha
\end{align}\]
In this question we used the triple angle formulas and trigonometric identities to find the value of angle so we must keep all these in mind and how to use it. With this we have to also remember the theorem of general solutions and trigonometric table of values of all the functions at each and every angle with their domain, range, period interval and all the important characteristics.