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If $\sin 3 \theta=\sin \theta$, how many solutions exist such that $-2 \pi < \theta < 2 \pi$ ?
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Verified Answer
The correct answer is:
5
We have, $\sin 3 \theta=\sin \theta$
$$
\begin{aligned}
&\Rightarrow \quad \sin 3 \theta-\sin \theta=0 \\
&\Rightarrow 2 \cos \left(\frac{3 \theta+\theta}{2}\right) \sin \left(\frac{3 \theta-\theta}{2}\right)=0 \\
&\Rightarrow \cos 2 \theta \cdot \sin \theta=0 \quad \cos 2 \theta=0 \quad \text { or } \sin \theta=0, \pi, 2 \pi \\
&\Rightarrow \quad \cos 2 \theta=\cos \left(\frac{\pi}{2}\right) \text { or } \quad \theta=0, \pi=2 \pi \\
&\Rightarrow \quad \quad \quad \quad \quad \quad \theta=0, \pi, 2 \pi \\
&\Rightarrow \quad \theta=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \quad \text { or } \quad \theta=0, \pi, 2 \pi \\
&\because \quad \theta=0, \pi, 2 \pi
\end{aligned}
$$
Thus, total number of solutions $=7$
$$
\begin{aligned}
&\Rightarrow \quad \sin 3 \theta-\sin \theta=0 \\
&\Rightarrow 2 \cos \left(\frac{3 \theta+\theta}{2}\right) \sin \left(\frac{3 \theta-\theta}{2}\right)=0 \\
&\Rightarrow \cos 2 \theta \cdot \sin \theta=0 \quad \cos 2 \theta=0 \quad \text { or } \sin \theta=0, \pi, 2 \pi \\
&\Rightarrow \quad \cos 2 \theta=\cos \left(\frac{\pi}{2}\right) \text { or } \quad \theta=0, \pi=2 \pi \\
&\Rightarrow \quad \quad \quad \quad \quad \quad \theta=0, \pi, 2 \pi \\
&\Rightarrow \quad \theta=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \quad \text { or } \quad \theta=0, \pi, 2 \pi \\
&\because \quad \theta=0, \pi, 2 \pi
\end{aligned}
$$
Thus, total number of solutions $=7$
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