The number of solutions of the equation sin A | vedclass

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(B) Given equation: $\sin A - 5 \sin 2A + \sin 3A = \cos A - 5 \cos 2A + \cos 3A$ Rearranging terms: $(\sin A + \sin 3A) - 5 \sin 2A = (\cos A + \cos

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$2$

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(B) Given equation: $\sin A - 5 \sin 2A + \sin 3A = \cos A - 5 \cos 2A + \cos 3A$ Rearranging terms: $(\sin A + \sin 3A) - 5 \sin 2A = (\cos A + \cos 3A) - 5 \cos 2A$ Using sum-to-product formulas: $2 \sin 2A \cos A - 5 \sin 2A = 2 \cos 2A \cos A - 5 \cos 2A$ Factoring: $2 \cos A (\sin 2A - \cos 2A) - 5 (\sin 2A - \cos 2A) = 0$ $(\sin 2A - \cos 2A)(2 \cos A - 5) = 0$ Since $2 \cos A - 5 = 0$ implies $\cos A = 2.5$,which is impossible,we must have $\sin 2A - \cos 2A = 0$ $\sin 2A = \cos 2A \Rightarrow 	an 2A = 1$ For $A \in (0, \pi)$,$2A \in (0, 2\pi)$ $	an 2A = 1$ at $2A = \frac{\pi}{4}$ and $2A = \frac{5\pi}{4}$ Thus,$A = \frac{\pi}{8}$ and $A = \frac{5\pi}{8}$ There are $2$ solutions in the given interval.

The number of solutions of the equation $\sin A - 5 \sin 2A + \sin 3A = \cos A - 5 \cos 2A + \cos 3A$ in the interval $(0, \pi)$ is:

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