Question

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

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

Answer 1

The correct option is A

$\frac{n\mathrm{\pi }}{4},\mathrm{n}\pi \pm \frac{\pi }{3}$

Explanation for the correct option.

Step 1: Simplify the expression

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

Use identity, $\sin a+\sin b=2\sin \left(\frac{a+b}{2}\right)\cos \left(\frac{a-b}{2}\right)$.

$\begin{array}{rcl}\left(\sin 6\theta +\sin 2\theta \right)+\sin 4\theta & =& 0\\ & \Rightarrow & 2\sin \left(\frac{6\theta +2\theta }{2}\right)\cos \left(\frac{6\theta -2\theta }{2}\right)+\sin 4\theta =0\\ & \Rightarrow & 2\sin 4\theta \cos 2\theta +\sin 4\theta =0\\ & \Rightarrow & \sin 4\theta \left(2\cos 2\theta +1\right)=0\end{array}$

Step 2: Find the solution

Using the zero property of the product we have:

$$\begin{gathered} \sin 4\theta =0,2\cos 2\theta +1=0 \\ \sin 4\theta =\sin n\pi ,\cos 2\theta =-\frac{1}{2} \end{gathered}$$

Comparing we have:

$$\begin{gathered} 4\theta =n\pi ,2\theta =2n\pi \pm \frac{2\pi }{3} \\ \theta =\frac{n\mathrm{\pi }}{4},\theta =n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3} \end{gathered}$$

So, the general solution of $\theta$ is $\frac{n\mathrm{\pi }}{4},\mathrm{n}\pi \pm \frac{\pi }{3}$.

Hence, option A is correct.