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  4. If sin 3 α=4 sin α sin (x+α) sin (x-α), then x=

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Q. If $\sin 3 \alpha=4 \sin \alpha \sin (x+\alpha) \sin (x-\alpha)$, then $x=$

Trigonometric Functions

Solution:

$\sin 3 \alpha=4 \sin \alpha \sin (x+\alpha) \sin (x-\alpha)$
$\therefore \sin 3 \alpha=4 \sin \alpha\left(\sin ^{2} x \cos ^{2} \alpha-\cos ^{2} x \sin ^{2} \alpha\right)$
$\therefore 3 \sin \alpha-4 \sin ^{3} \alpha=4 \sin \alpha\left(\sin ^{2} x-\sin ^{2} \alpha\right)$
$\therefore \sin ^{2} x=\left(\frac{3}{4}\right) \Rightarrow \sin ^{2} x=\sin ^{2} \frac{\pi}{3} \,\,\,\therefore x= n \pi \pm \frac{\pi}{3}$