Question

The general solution to the equation $\sin 3 \alpha=4 \sin \alpha\sin (x+\alpha) \sin (x-\alpha)$ is

• A. $n \pi \pm \pi / 4, \forall n \in I$
• B. $n \pi \pm \pi / 3, \forall n \in I$
• C. $n \pi \pm \pi / 9, \forall n \in I$
• D. $n \pi \pm \pi / 12, \forall n \in I$

Answer 1

Answer: (B)

We have $\sin 3 \alpha=4 \sin \alpha\left(\sin ^{2} x-\sin ^{2} \alpha\right)$
$\Rightarrow 3 \sin \alpha-4 \sin ^{3} \alpha=4 \sin \alpha \sin ^{2} x-4 \sin ^{3} \alpha$
$\Rightarrow 3 \sin \alpha=4 \sin \alpha \sin ^{2} x$
or $\sin \alpha=0$
If $\sin \alpha \neq 0, \sin ^{2} x=3 / 4=(\sqrt{3} / 2)^{2}=\sin ^{2}(\pi / 3)$
$\therefore x=n \pi \pm \pi / 3, \forall n \in Z$
If $\sin \alpha=0$, i.e., $\alpha=n \pi$, equation becomes an identity.