Question

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

• A. $\Rightarrow$
• B. $\sin 4\theta(2\cos 2\theta + 1) = 0$
• C. $\Rightarrow$
• D. $2\cos 2\theta = - 1$

Answer 1

Answer: (A)

$\Rightarrow$

Answer 2

$\Rightarrow$

$\sin\left( \frac{\pi}{4} + 2\theta \right) + \cos\left( \frac{\pi}{4} + 2\theta \right) = \frac{1}{\sqrt{2}} \Rightarrow$

$\cos 2\theta = \frac{1}{2} \Rightarrow 2\theta = 2n\pi \pm \frac{\pi}{3} \Rightarrow \theta = n\pi \pm \frac{\pi}{6}$ $( \tan \theta - \sqrt { 3 } ) ( \tan \theta - 1 ) = 0$ $\cos\theta = \frac{- 3 \pm \sqrt{9 + 8}}{4} = \frac{- 3 \pm \sqrt{17}}{4} \Rightarrow$, $\theta = 2n\pi \pm \cos^{- 1}\left( \frac{- 3 + \sqrt{17}}{4} \right)$.