The number of possible solutions of sin theta | vedclass

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(C) Given equation: $\sin 7 	heta + \sin 	heta + \sin 4 	heta = 0$ Using the sum-to-product formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \f

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$6$

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(C) Given equation: $\sin 7 	heta + \sin 	heta + \sin 4 	heta = 0$ Using the sum-to-product formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$: $2 \sin 4 	heta \cos 3 	heta + \sin 4 	heta = 0$ $\sin 4 	heta (2 \cos 3 	heta + 1) = 0$ Case $1$: $\sin 4 	heta = 0 \implies 4 	heta = n \pi \implies 	heta = \frac{n \pi}{4}$. For $	heta \in (0, \pi)$,$n = 1, 2, 3$,so $	heta = \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}$. Case $2$: $\cos 3 	heta = -\frac{1}{2} = \cos \frac{2 \pi}{3} \implies 3 	heta = 2 n \pi \pm \frac{2 \pi}{3} \implies 	heta = \frac{2 n \pi}{3} \pm \frac{2 \pi}{9}$. For $	heta \in (0, \pi)$: If $n=0$,$	heta = \frac{2 \pi}{9}$ (since $-\frac{2 \pi}{9}$ is outside range). If $n=1$,$	heta = \frac{2 \pi}{3} \pm \frac{2 \pi}{9} = \frac{8 \pi}{9}, \frac{4 \pi}{9}$. If $n=2$,$	heta = \frac{4 \pi}{3} \pm \frac{2 \pi}{9}$ (both outside range). The solutions are $	heta \in \{\frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{2 \pi}{9}, \frac{4 \pi}{9}, \frac{8 \pi}{9}\}$. Total number of solutions $= 6$.

The number of possible solutions of $\sin \theta + \sin 4 \theta + \sin 7 \theta = 0$ for $\theta \in (0, \pi)$ is:

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