A value of theta satisfying sin 5theta - sin | vedclass

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(B) Given equation: $\sin 5	heta - \sin 3	heta + \sin 	heta = 0$ for $	heta \in (0, \frac{\pi}{2})$.Rearranging the terms: $(\sin 5	heta + \sin \

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$\frac{\pi}{6}$

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(B) Given equation: $\sin 5	heta - \sin 3	heta + \sin 	heta = 0$ for $	heta \in (0, \frac{\pi}{2})$.Rearranging the terms: $(\sin 5	heta + \sin 	heta) = \sin 3	heta$.Using the sum-to-product formula $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:$2 \sin 3	heta \cos 2	heta = \sin 3	heta$.$2 \sin 3	heta \cos 2	heta - \sin 3	heta = 0$.$\sin 3	heta (2 \cos 2	heta - 1) = 0$.This gives two cases:Case $1$: $\sin 3	heta = 0$ $\Rightarrow 3	heta = n\pi$ $\Rightarrow 	heta = \frac{n\pi}{3}$. For $0 Case $2$: $2 \cos 2	heta - 1 = 0 \Rightarrow \cos 2	heta = \frac{1}{2} = \cos \frac{\pi}{3}$.$2	heta = 2n\pi \pm \frac{\pi}{3} \Rightarrow 	heta = n\pi \pm \frac{\pi}{6}$. For $0 Thus,the values of $	heta$ are $\frac{\pi}{6}$ and $\frac{\pi}{3}$. Comparing with the options,$\frac{\pi}{6}$ is the correct choice.

$A$ value of $\theta$ satisfying $\sin 5\theta - \sin 3\theta + \sin \theta = 0$,such that $0 < \theta < \frac{\pi}{2}$ is

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