Question

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

• A. $\frac{n\pi }{4}$, n$\pi$ $\frac{\pi }{3}$
• B. $\frac{n\pi }{4}$, n$\pi$ $\frac{\pi }{6}$
• C. $\frac{n\pi }{4}$, 2n$\pi$ $\frac{\pi }{3}$
• D. $\frac{n\pi }{4}$, 2n$\pi$ $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{n\pi }{4}$, n$\pi$ $\frac{\pi }{6}$

Given $\sin 6\theta +\sin 4\theta +\sin 2\theta =0$

$⟹2\sin 3\theta \cos 3\theta +2\sin 3\theta \cos \theta =0$ $(\because \sin A+\sin B=2\sin \frac{A+B}{2}\cos \frac{A-B}{2},\sin 2A=2\sin A\cos A)$

$\sin 3\theta (\cos 3\theta +\cos \theta )=0$

$⟹2\sin 3\theta \cos 2\theta \cos \theta =0$

$⟹\sin 3\theta =0$ or $\cos 2\theta =0$ or $\cos \theta =0$

$⟹\theta =\frac{n\pi }{3},\frac{(2n+1)}{4},\frac{(2n+1)\pi }{2}$