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The general solution of the equation $3 \sec ^2 \theta=2 \operatorname{cosec} \theta$ is
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Verified Answer
The correct answer is:
$\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{6}, \mathrm{n} \in \mathrm{Z}$
$$
\begin{aligned}
& 3 \sec ^2 \theta=2 \operatorname{cosec} \theta \\
& \Rightarrow \frac{3}{\cos ^2 \theta}=\frac{2}{\sin \theta} \\
& \Rightarrow \frac{3}{1-\sin ^2 \theta}=\frac{2}{\sin \theta} \\
& \Rightarrow 2 \sin ^2 \theta+3 \sin \theta-2=0 \\
& \Rightarrow(2 \sin \theta-1)(\sin \theta+2)=0 \\
& \Rightarrow \sin \theta=\frac{1}{2}
\end{aligned}
$$
or $\sin \theta=-2$, which is not possible
$$
\begin{aligned}
\therefore \quad \sin \theta & =\frac{1}{2}=\sin \frac{\pi}{6} \\
\Rightarrow \theta & =\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{6}, \mathrm{n} \in Z
\end{aligned}
$$
\begin{aligned}
& 3 \sec ^2 \theta=2 \operatorname{cosec} \theta \\
& \Rightarrow \frac{3}{\cos ^2 \theta}=\frac{2}{\sin \theta} \\
& \Rightarrow \frac{3}{1-\sin ^2 \theta}=\frac{2}{\sin \theta} \\
& \Rightarrow 2 \sin ^2 \theta+3 \sin \theta-2=0 \\
& \Rightarrow(2 \sin \theta-1)(\sin \theta+2)=0 \\
& \Rightarrow \sin \theta=\frac{1}{2}
\end{aligned}
$$
or $\sin \theta=-2$, which is not possible
$$
\begin{aligned}
\therefore \quad \sin \theta & =\frac{1}{2}=\sin \frac{\pi}{6} \\
\Rightarrow \theta & =\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{6}, \mathrm{n} \in Z
\end{aligned}
$$
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