If sin(x+3alpha) + 3sin(x-alpha) = 0,then | vedclass

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(C) Given equation: $\sin(x+3\alpha) + 3\sin(x-\alpha) = 0$Expanding using $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$:$(\sin x \cos 3\alpha + \

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$	an x = 	an^3 \alpha$

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(C) Given equation: $\sin(x+3\alpha) + 3\sin(x-\alpha) = 0$Expanding using $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$:$(\sin x \cos 3\alpha + \cos x \sin 3\alpha) + 3(\sin x \cos \alpha - \cos x \sin \alpha) = 0$Using $\cos 3\alpha = 4\cos^3 \alpha - 3\cos \alpha$ and $\sin 3\alpha = 3\sin \alpha - 4\sin^3 \alpha$:$\sin x(4\cos^3 \alpha - 3\cos \alpha) + \cos x(3\sin \alpha - 4\sin^3 \alpha) + 3\sin x \cos \alpha - 3\cos x \sin \alpha = 0$$\sin x(4\cos^3 \alpha - 3\cos \alpha + 3\cos \alpha) + \cos x(3\sin \alpha - 4\sin^3 \alpha - 3\sin \alpha) = 0$$4\sin x \cos^3 \alpha - 4\cos x \sin^3 \alpha = 0$$4\sin x \cos^3 \alpha = 4\cos x \sin^3 \alpha$$\frac{\sin x}{\cos x} = \frac{\sin^3 \alpha}{\cos^3 \alpha}$$	an x = 	an^3 \alpha$

If $\sin(x+3\alpha) + 3\sin(x-\alpha) = 0$,then

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