If sin 3alpha = 4sin alpha sin (x + alpha )si | vedclass

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(B) Given equation: $\sin 3\alpha = 4\sin \alpha \sin (x + \alpha )\sin (x - \alpha )$Using the identity $\sin (x + \alpha )\sin (x - \alpha ) = \sin^

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$n\pi \pm \frac{\pi }{3}$

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(B) Given equation: $\sin 3\alpha = 4\sin \alpha \sin (x + \alpha )\sin (x - \alpha )$Using the identity $\sin (x + \alpha )\sin (x - \alpha ) = \sin^2 x - \sin^2 \alpha$,we get:$\sin 3\alpha = 4\sin \alpha (\sin^2 x - \sin^2 \alpha)$Using the identity $\sin 3\alpha = 3\sin \alpha - 4\sin^3 \alpha$:$3\sin \alpha - 4\sin^3 \alpha = 4\sin \alpha \sin^2 x - 4\sin \alpha \sin^2 \alpha$$3\sin \alpha - 4\sin^3 \alpha = 4\sin \alpha \sin^2 x - 4\sin^3 \alpha$$3\sin \alpha = 4\sin \alpha \sin^2 x$Assuming $\sin \alpha 
eq 0$,we divide by $\sin \alpha$:$3 = 4\sin^2 x$$\sin^2 x = \frac{3}{4} = \left(\frac{\sqrt{3}}{2}ight)^2$$\sin^2 x = \sin^2 \left(\frac{\pi}{3}ight)$The general solution for $\sin^2 x = \sin^2 	heta$ is $x = n\pi \pm 	heta$.Therefore,$x = n\pi \pm \frac{\pi}{3}$.

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

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