MCQ
જો $\sin 3\alpha = 4\sin \alpha \sin (x + \alpha )\sin (x - \alpha ),$ તો $x = $
- A$n\pi \pm \frac{\pi }{6}$
- ✓$n\pi \pm \frac{\pi }{3}$
- C$n\pi \pm \frac{\pi }{4}$
- D$n\pi \pm \frac{\pi }{2}$
$\therefore $ ${\sin ^2}x = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2}$
==> ${\sin ^2}x = {\sin ^2}\pi /3$
==> $x = n\pi \pm \pi /3$.
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