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Question

Mathematics Question on Trigonometric Functions

If sin6θ+sin4θ+sin2θ=0\sin 6\theta + \sin 4\theta + \sin 2\theta = 0, then general value of θ\theta is

A

nπ4,nπ±π3\frac{n\pi}{4},n\pi\pm\frac{\pi}{3}

B

nπ4,nπ±π4\frac{n\pi}{4},n\pi\pm\frac{\pi}{4}

C

nπ4,nπ±π6\frac{n\pi}{4},n\pi\pm\frac{\pi}{6}

D

nπ4,2nπ±π6\frac{n\pi}{4},2n\pi\pm\frac{\pi}{6}

Answer

nπ4,nπ±π3\frac{n\pi}{4},n\pi\pm\frac{\pi}{3}

Explanation

Solution

sin6θ+sin4θ+sin2θ=0\sin 6\theta + \sin 4\theta + \sin 2\theta = 0 (sin6θ+sin2θ)+sin4θ=0\Rightarrow\, \, \, \, \, \, (\sin 6\theta + \sin \: 2\theta) + \sin 4\theta = 0 2sin4θcos2θ+sin4θ=0\Rightarrow\, \, \, \, \, 2 \sin 4\theta \cos 2\theta + \sin 4\theta = 0 sin4θ(2cos2θ+1)=0\Rightarrow \, \, \, \, \, \, \sin 4\theta (2 \cos 2\theta + 1) = 0 sin4θ=0\Rightarrow\, \, \, \, \, \, \, \, \sin 4\theta = 0 or cos2θ=12=cos2π3\cos 2\theta = -\frac{1}{2} = \cos \frac{2 \pi}{3} 4θ=nπ\Rightarrow \, \, \, \, \, \, 4\theta = n \pi or 2θ=2nπ±232 \theta = 2 n \pi \pm \frac{2}{3} θ=nπ4,θ=nπ±π3θ\Rightarrow \, \, \, \, \, \, \theta = \frac{n \pi}{4} , \theta = n \pi \pm \frac{\pi}{3} \, \theta \,