The number of solutions of sin x + sin 3x + s | vedclass

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(B) Given equation: $\sin x + \sin 3x + \sin 5x = 0$ Grouping terms: $(\sin 5x + \sin x) + \sin 3x = 0$ Using $\sin C + \sin D = 2 \sin \left(\frac{C+

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$3$

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(B) Given equation: $\sin x + \sin 3x + \sin 5x = 0$ Grouping terms: $(\sin 5x + \sin x) + \sin 3x = 0$ Using $\sin C + \sin D = 2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$: $2 \sin 3x \cos 2x + \sin 3x = 0$ $\sin 3x (2 \cos 2x + 1) = 0$ This implies $\sin 3x = 0$ or $\cos 2x = -\frac{1}{2}$. Case $1$: $\sin 3x = 0 \implies 3x = n\pi \implies x = \frac{n\pi}{3}$. For $x \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$,possible values are $x = \frac{2\pi}{3}, \pi, \frac{4\pi}{3}$. Case $2$: $\cos 2x = -\frac{1}{2} \implies 2x = 2n\pi \pm \frac{2\pi}{3} \implies x = n\pi \pm \frac{\pi}{3}$. For $x \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$,possible values are $x = \frac{2\pi}{3}, \frac{4\pi}{3}$. Combining both cases,the distinct solutions are $x = \frac{2\pi}{3}, \pi, \frac{4\pi}{3}$. Thus,the total number of solutions is $3$.

The number of solutions of $\sin x + \sin 3x + \sin 5x = 0$ in the interval $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ is

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