Question

If $sin3\theta =Sin\theta$ , how many solutions exist such that $-2\pi <\theta <2\pi$?

• A. $9$
• B. $8$
• C. $7$
• D. $5$

Answer 1

Answer: (C)

We have, $\sin 3\theta =\sin \theta$
$\Rightarrow \sin 3\theta -\sin \theta =0$
$\Rightarrow 2\cos \left(\frac{3\theta +\theta }{2}\right)\sin \left(\frac{3\theta -\theta }{2}\right)=0$
$\Rightarrow \cos 2\theta \cdot \sin \theta =0$
$\Rightarrow \cos 2\theta =0$
$\Rightarrow \cos 2\theta =\cos \left(\frac{\pi }{2}\right)\text{ or }\sin \theta =0,\pi ,2\pi$
or $\theta =0,\pi =2\pi$
$\Rightarrow 2\theta =\frac{\pi }{2}$ or $\theta =0,\pi ,2\pi$
$\Rightarrow \theta =\frac{n\pi }{4}$ or $\theta =0,\pi ,2\pi$
$-2\pi <\theta <2\pi$
$\theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ or $\theta =0,\pi ,2\pi$
Thus, total number of solutions =7