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

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(B) 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)

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$\frac{n\pi}{2} + \frac{\pi}{8}$

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(B) 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 $2\cos x - 3 
eq 0$ (as $\cos x$ cannot be $1.5$),we have $\sin 2x = \cos 2x$$	an 2x = 1$$2x = n\pi + \frac{\pi}{4}$$x = \frac{n\pi}{2} + \frac{\pi}{8}$

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

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