The general solution of sin x + sin 5x = sin | vedclass

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(C) Given equation: $\sin x + \sin 5x = \sin 2x + \sin 4x$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2}

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$n\pi / 3$

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(C) Given equation: $\sin x + \sin 5x = \sin 2x + \sin 4x$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$2 \sin(3x) \cos(-2x) = 2 \sin(3x) \cos(-x)$Since $\cos(-	heta) = \cos 	heta$,we have:$2 \sin(3x) \cos(2x) = 2 \sin(3x) \cos(x)$$2 \sin(3x) [\cos(2x) - \cos(x)] = 0$Case $1$: $\sin(3x) = 0 \implies 3x = n\pi \implies x = \frac{n\pi}{3}$Case $2$: $\cos(2x) - \cos(x) = 0 \implies \cos(2x) = \cos(x)$$2x = 2n\pi \pm x$For $+$,$x = 2n\pi$. For $-$,$3x = 2n\pi \implies x = \frac{2n\pi}{3}$Since $\frac{n\pi}{3}$ covers all solutions including $\frac{2n\pi}{3}$ and $2n\pi$,the general solution is $x = \frac{n\pi}{3}$.

The general solution of $\sin x + \sin 5x = \sin 2x + \sin 4x$ is:

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