If sin 2x + sin 4x = 2sin 3x,then x = | vedclass

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(A) Given equation: $\sin 2x + \sin 4x = 2\sin 3x$Using the sum-to-product formula $\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$:$2\sin(

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

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(A) Given equation: $\sin 2x + \sin 4x = 2\sin 3x$Using the sum-to-product formula $\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$:$2\sin(\frac{2x+4x}{2})\cos(\frac{2x-4x}{2}) = 2\sin 3x$$2\sin 3x \cos(-x) = 2\sin 3x$Since $\cos(-x) = \cos x$,we have:$2\sin 3x \cos x - 2\sin 3x = 0$$2\sin 3x(\cos x - 1) = 0$This implies either $\sin 3x = 0$ or $\cos x = 1$.If $\sin 3x = 0$,then $3x = n\pi \Rightarrow x = \frac{n\pi}{3}$.If $\cos x = 1$,then $x = 2n\pi$.Since $2n\pi$ is a subset of $\frac{n\pi}{3}$ (when $n$ is a multiple of $6$),the general solution is $x = \frac{n\pi}{3}$.

If $\sin 2x + \sin 4x = 2\sin 3x$,then $x =$

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