Find the general solution of sin x + sin 2x + | vedclass

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

Vedclass · Accepted Answer

$2n\pi \pm \frac{2\pi}{3}, \frac{n\pi}{2} + \frac{\pi}{8}, n \in Z$

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(D) Given equation: $\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x$ Grouping terms: $(\sin 3x + \sin x) + \sin 2x = (\cos 3x + \cos x) + \cos 2x$ Using sum-to-product formulas: $2\sin 2x \cos x + \sin 2x = 2\cos 2x \cos x + \cos 2x$ Factoring out common terms: $\sin 2x(2\cos x + 1) = \cos 2x(2\cos x + 1)$ Rearranging: $(\sin 2x - \cos 2x)(2\cos x + 1) = 0$ Case $1$: $\sin 2x - \cos 2x = 0 \implies 	an 2x = 1 = 	an \frac{\pi}{4}$ $2x = n\pi + \frac{\pi}{4} \implies x = \frac{n\pi}{2} + \frac{\pi}{8}$ Case $2$: $2\cos x + 1 = 0 \implies \cos x = -\frac{1}{2} = \cos \frac{2\pi}{3}$ $x = 2n\pi \pm \frac{2\pi}{3}$ Combining both,the general solution is $x = 2n\pi \pm \frac{2\pi}{3}$ and $x = \frac{n\pi}{2} + \frac{\pi}{8}$ for $n \in Z$. Hence,option $(D)$ is correct.

Find the general solution of $\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x$.

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