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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:
9
We have,
$$
\begin{aligned}
&\Rightarrow \quad \sin 3 \theta=\sin \theta \\
&\Rightarrow \quad \quad 3 \sin \theta-4 \sin ^{3} \theta=\sin \theta \\
&\Rightarrow \quad \quad 4 \sin ^{3} \theta-2 \sin \theta=0 \\
&\Rightarrow \quad 2 \sin \theta\left(2 \sin ^{2} \theta-1\right)=0 \\
&\Rightarrow \quad \sin \theta=0 \text { or } \sin \theta=\pm \frac{1}{\sqrt{2}} \\
&\text { If } \sin \theta=0 \text { and if } \sin \theta=\pm \frac{1}{\sqrt{2}} \\
&\Rightarrow \quad \theta=0, \pi,-\pi \\
&\Rightarrow \quad \theta, \frac{3 \pi}{4}, \frac{-5 \pi}{4}, \frac{-7 \pi}{4}, \frac{-\pi}{4}, \frac{-3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}
\end{aligned}
$$
$\therefore$ There are nine solutions.
$$
\begin{aligned}
&\Rightarrow \quad \sin 3 \theta=\sin \theta \\
&\Rightarrow \quad \quad 3 \sin \theta-4 \sin ^{3} \theta=\sin \theta \\
&\Rightarrow \quad \quad 4 \sin ^{3} \theta-2 \sin \theta=0 \\
&\Rightarrow \quad 2 \sin \theta\left(2 \sin ^{2} \theta-1\right)=0 \\
&\Rightarrow \quad \sin \theta=0 \text { or } \sin \theta=\pm \frac{1}{\sqrt{2}} \\
&\text { If } \sin \theta=0 \text { and if } \sin \theta=\pm \frac{1}{\sqrt{2}} \\
&\Rightarrow \quad \theta=0, \pi,-\pi \\
&\Rightarrow \quad \theta, \frac{3 \pi}{4}, \frac{-5 \pi}{4}, \frac{-7 \pi}{4}, \frac{-\pi}{4}, \frac{-3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}
\end{aligned}
$$
$\therefore$ There are nine solutions.
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