If sin 6 theta + sin 4 theta + sin 2 theta = | vedclass

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

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

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(A) Given equation: $\sin 6 	heta + \sin 4 	heta + \sin 2 	heta = 0$Grouping terms: $(\sin 6 	heta + \sin 2 	heta) + \sin 4 	heta = 0$Using the formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$:$2 \sin 4 	heta \cos 2 	heta + \sin 4 	heta = 0$Factoring out $\sin 4 	heta$:$\sin 4 	heta (2 \cos 2 	heta + 1) = 0$This gives two cases:Case $1$: $\sin 4 	heta = 0$ $\Rightarrow 4 	heta = n \pi$ $\Rightarrow 	heta = \frac{n \pi}{4}$Case $2$: $2 \cos 2 	heta + 1 = 0 \Rightarrow \cos 2 	heta = -\frac{1}{2} = \cos \frac{2 \pi}{3}$General solution for $\cos x = \cos \alpha$ is $x = 2 n \pi \pm \alpha$:$2 	heta = 2 n \pi \pm \frac{2 \pi}{3} \Rightarrow 	heta = n \pi \pm \frac{\pi}{3}$Thus, $	heta = \frac{n \pi}{4}$ or $	heta = n \pi \pm \frac{\pi}{3}$, where $n \in \mathbb{Z}$.

If $\sin 6 \theta + \sin 4 \theta + \sin 2 \theta = 0$, then the general value of $\theta$ is

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