Question

solve sin 2 theta + sin 4 theta + sin 6 theta =0

Answer 1

mpossible it is  12 theta

Answer 2

$\bold{\theta = \frac{(2n + 1)\times \pi}{2} \pm \frac{\pi}{6}}$

Step-by-step explanation:

Given,

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

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

∵ $sin A + sin B = 2\sin \frac{(A + B)}{2} \times cos \frac{(A - B)}{2}$

∴ $2 \sin 4\theta \times cos 2 \theta + sin 4\theta = 0$

⇒ $sin 4 \theta\times (2\times cos 2\theta + 1) = 0$

$sin 4\theta = 0$ and , $2 \times cos 2 \theta + 1 = 0$

From,$sin 4\theta = 0$

$\theta = \frac{n\times \pi}{4}$

From $2 cos 2\theta + 1 = 0$ ,

$2\theta = (2n + 1)\times \pi \pm \frac{\pi}{3}$

$\bold{\theta = \frac{(2n + 1)\times \pi}{2} \pm \frac{\pi}{6}}$