If sin 6theta + sin 4theta + sin 2theta = 0,t | vedclass

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Question

(A) Given equation: $\sin 6	heta + \sin 4	heta + \sin 2	heta = 0$Group the terms: $(\sin 6	heta + \sin 2	heta) + \sin 4	heta = 0$Using the formu

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$\frac{n\pi}{4}$ or $n\pi \pm \frac{\pi}{3}$

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(A) Given equation: $\sin 6	heta + \sin 4	heta + \sin 2	heta = 0$Group the terms: $(\sin 6	heta + \sin 2	heta) + \sin 4	heta = 0$Using the formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$:$2 \sin 4	heta \cos 2	heta + \sin 4	heta = 0$Factor out $\sin 4	heta$:$\sin 4	heta (2 \cos 2	heta + 1) = 0$Case $1$: $\sin 4	heta = 0$$4	heta = n\pi \Rightarrow 	heta = \frac{n\pi}{4}$Case $2$: $2 \cos 2	heta + 1 = 0$$\cos 2	heta = -\frac{1}{2}$Since $\cos \frac{2\pi}{3} = -\frac{1}{2}$,the general solution is:$2	heta = 2n\pi \pm \frac{2\pi}{3}$$	heta = n\pi \pm \frac{\pi}{3}$Thus,$	heta = \frac{n\pi}{4}$ or $	heta = n\pi \pm \frac{\pi}{3}$.

If $\sin 6\theta + \sin 4\theta + \sin 2\theta = 0$,then $\theta = $

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