If sin 3alpha = 4 sin alpha cdot sin(x + alph | vedclass

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

Vedclass · Accepted Answer

$x = n\pi \pm \frac{\pi}{3}, n \in Z$

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(A) Given equation: $\sin 3\alpha = 4 \sin \alpha \cdot \sin(x + \alpha) \cdot \sin(x - \alpha)$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 $\sin 3\alpha = 3 \sin \alpha - 4 \sin^3 \alpha$, we get:$3 \sin \alpha - 4 \sin^3 \alpha = 4 \sin \alpha \sin^2 x - 4 \sin \alpha \sin^2 \alpha$Since $\alpha 
eq n\pi$, $\sin \alpha 
eq 0$, we can divide by $\sin \alpha$:$3 - 4 \sin^2 \alpha = 4 \sin^2 x - 4 \sin^2 \alpha$$3 = 4 \sin^2 x$$\sin^2 x = \frac{3}{4} = (\frac{\sqrt{3}}{2})^2 = \sin^2(\frac{\pi}{3})$The general solution for $\sin^2 x = \sin^2 	heta$ is $x = n\pi \pm 	heta$.Therefore, $x = n\pi \pm \frac{\pi}{3}, n \in Z$.

If $\sin 3\alpha = 4 \sin \alpha \cdot \sin(x + \alpha) \cdot \sin(x - \alpha)$ where $\alpha \neq n\pi, n \in Z$, then all possible values of $x$ are given as

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