The number of solutions of the equation sin 7 | vedclass

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(D) Given equation: $\sin 7 	heta - \sin 3 	heta = \sin 4 	heta$ Using the identity $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$,we

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$5$

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(D) Given equation: $\sin 7 	heta - \sin 3 	heta = \sin 4 	heta$ Using the identity $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$,we get: $2 \cos 5 	heta \sin 2 	heta = \sin 4 	heta$ Since $\sin 4 	heta = 2 \sin 2 	heta \cos 2 	heta$,the equation becomes: $2 \cos 5 	heta \sin 2 	heta = 2 \sin 2 	heta \cos 2 	heta$ $2 \sin 2 	heta (\cos 5 	heta - \cos 2 	heta) = 0$ Case $1$: $\sin 2 	heta = 0$ $2 	heta = n \pi \Rightarrow 	heta = \frac{n \pi}{2}$. For $	heta \in (0, \pi)$,$	heta = \frac{\pi}{2}$. Case $2$: $\cos 5 	heta = \cos 2 	heta$ $5 	heta = 2 n \pi \pm 2 	heta$ If $5 	heta = 2 n \pi + 2 	heta$,then $3 	heta = 2 n \pi \Rightarrow 	heta = \frac{2 n \pi}{3}$. For $	heta \in (0, \pi)$,$	heta = \frac{2 \pi}{3}$. If $5 	heta = 2 n \pi - 2 	heta$,then $7 	heta = 2 n \pi \Rightarrow 	heta = \frac{2 n \pi}{7}$. For $	heta \in (0, \pi)$,$	heta = \frac{2 \pi}{7}, \frac{4 \pi}{7}, \frac{6 \pi}{7}$. The solutions are $\frac{\pi}{2}, \frac{2 \pi}{3}, \frac{2 \pi}{7}, \frac{4 \pi}{7}, \frac{6 \pi}{7}$. Total number of solutions is $5$.

The number of solutions of the equation $\sin 7 \theta - \sin 3 \theta = \sin 4 \theta$ that lie in the interval $(0, \pi)$ is

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