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Trigonometrical Equations
medium
જો $\sin 3\alpha = 4\sin \alpha \sin (x + \alpha )\sin (x - \alpha ),$ તો $x = $
A
$n\pi \pm \frac{\pi }{6}$
B
$n\pi \pm \frac{\pi }{3}$
C
$n\pi \pm \frac{\pi }{4}$
D
$n\pi \pm \frac{\pi }{2}$
Solution
(b) $3\sin \alpha – 4{\sin ^3}\alpha = 4\sin \alpha ({\sin ^2}x – {\sin ^2}\alpha )$
$\therefore $ ${\sin ^2}x = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2}$
==> ${\sin ^2}x = {\sin ^2}\pi /3$
==> $x = n\pi \pm \pi /3$.
Standard 11
Mathematics
