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 $\sin \alpha \ne 0,{\sin }^{2}x=3/4=(\sqrt{3}/2{)}^2={\sin }^2(\pi /3)$
$\therefore x=n\pi \pm \pi /3,\forall n\in Z$
If $\sin \alpha =0$, i.e., $\alpha =n\pi$, equation becomes an identity.