The general solution of sin x - 3 sin 2x + si | vedclass

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(D) Given equation: $\sin x - 3 \sin 2x + \sin 3x = \cos x - 3 \cos 2x + \cos 3x$ Rearranging terms: $(\sin 3x + \sin x) - 3 \sin 2x = (\cos 3x + \cos

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$x = \frac{n\pi}{2} + \frac{\pi}{8}, n \in Z$

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(D) Given equation: $\sin x - 3 \sin 2x + \sin 3x = \cos x - 3 \cos 2x + \cos 3x$ Rearranging terms: $(\sin 3x + \sin x) - 3 \sin 2x = (\cos 3x + \cos x) - 3 \cos 2x$ Using sum-to-product formulas: $2 \sin 2x \cos x - 3 \sin 2x = 2 \cos 2x \cos x - 3 \cos 2x$ $\sin 2x (2 \cos x - 3) = \cos 2x (2 \cos x - 3)$ $(\sin 2x - \cos 2x)(2 \cos x - 3) = 0$ Since $2 \cos x - 3 = 0$ implies $\cos x = 1.5$,which is impossible,we have $\sin 2x = \cos 2x$ $	an 2x = 1$ $2x = n\pi + \frac{\pi}{4}$ $x = \frac{n\pi}{2} + \frac{\pi}{8}, n \in Z$

The general solution of $\sin x - 3 \sin 2x + \sin 3x = \cos x - 3 \cos 2x + \cos 3x$ is

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