The solutions of sin x + sin 5x = sin 3x in t | vedclass

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(B) Given equation: $\sin x + \sin 5x = \sin 3x$ Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$: $2 \s

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$\frac{\pi}{6}, \frac{\pi}{3}$

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(B) Given equation: $\sin x + \sin 5x = \sin 3x$ Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$: $2 \sin(\frac{x+5x}{2}) \cos(\frac{x-5x}{2}) = \sin 3x$ $2 \sin 3x \cos(-2x) = \sin 3x$ Since $\cos(-2x) = \cos 2x$,we have: $2 \sin 3x \cos 2x - \sin 3x = 0$ $\sin 3x (2 \cos 2x - 1) = 0$ Case $1$: $\sin 3x = 0 \implies 3x = n\pi \implies x = \frac{n\pi}{3}$. For $x \in (0, \frac{\pi}{2})$,$x = \frac{\pi}{3}$ is a solution. Case $2$: $2 \cos 2x - 1 = 0 \implies \cos 2x = \frac{1}{2} = \cos \frac{\pi}{3}$. $2x = 2n\pi \pm \frac{\pi}{3} \implies x = n\pi \pm \frac{\pi}{6}$. For $x \in (0, \frac{\pi}{2})$,$x = \frac{\pi}{6}$ is a solution. Thus,the solutions are $x = \frac{\pi}{6}, \frac{\pi}{3}$.

The solutions of $\sin x + \sin 5x = \sin 3x$ in the interval $(0, \frac{\pi}{2})$ are

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