Question

If $sin \, 3 \,\theta\, = \, Sin \, \theta$ , 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