If sin⁡6θ+sin⁡4θ+sin⁡2θ=0 then θ is equal of (n∈Z)

sin⁡6θ+sin⁡4θ+sin⁡2θ=0⇒ (sin⁡6θ+sin⁡2θ)+sin⁡4θ=0⇒ 2sin⁡4θcos⁡2θ+sin⁡4θ=0⇒ sin⁡4θ(2cos⁡2θ+1)=0⇒ sin⁡4θ=0 or cos⁡2θ=−12=cos⁡2π3⇒ 4θ=nπ or 2θ=2nπ±2π3,n∈Z⇒ θ=nπ4 or θ=nπ±π3

If sin⁡6θ+sin⁡4θ+sin⁡2θ=0 then θ is equal of (n∈Z)

sin⁡6θ+sin⁡4θ+sin⁡2θ=0⇒ (sin⁡6θ+sin⁡2θ)+sin⁡4θ=0⇒ 2sin⁡4θcos⁡2θ+sin⁡4θ=0⇒ sin⁡4θ(2cos⁡2θ+1)=0⇒ sin⁡4θ=0 or cos⁡2θ=−12=cos⁡2π3⇒ 4θ=nπ or 2θ=2nπ±2π3,n∈Z⇒ θ=nπ4 or θ=nπ±π3

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If $\sin 6 θ + \sin 4 θ + \sin 2 θ = 0$ then $θ$ is equal of $(n \in Z)$

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$\frac{n π}{4} or 2 n π \pm \frac{π}{6}$

none of these

$\frac{n π}{4} or n π \pm \frac{π}{6}$

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

answer is A.

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Detailed Solution

$\begin{array}{l} \sin 6 θ + \sin 4 θ + \sin 2 θ = 0 \\ \Rightarrow (\sin 6 θ + \sin 2 θ) + \sin 4 θ = 0 \\ \Rightarrow 2 \sin 4 θcos 2 θ + \sin 4 θ = 0 \\ \Rightarrow \sin 4 θ (2 \cos 2 θ + 1) = 0 \\ \Rightarrow \sin 4 θ = 0 or \cos 2 θ = - \frac{1}{2} = \cos \frac{2 π}{3} \\ \Rightarrow 4 θ = nπ or 2 θ = 2 nπ \pm \frac{2 π}{3}, n \in Z \\ \Rightarrow θ = \frac{nπ}{4} or θ = nπ \pm \frac{π}{3} \end{array}$