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The set of values of $\theta$ satisfying the inequation $2 \sin ^2 \theta-5 \sin \theta+2>0$, where $0 < \theta < 2 \pi$, is
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Verified Answer
The correct answer is:
$\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, 2 \pi\right)$
$\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, 2 \pi\right)$
$$
\begin{aligned}
& \text { As, } 2 \sin ^2 \theta-5 \sin \theta+2>0 \\
& \Rightarrow(2 \sin \theta-1)(\sin \theta-2)>0 \\
& \quad[\text { where, }(\sin \theta-2) < 0 \text { for all } \theta \in R] \\
& \therefore \quad(2 \sin \theta-1) < 0
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & \sin \theta < \frac{1}{2}, \text { shown as } \\
\therefore & \theta \in\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, 2 \pi\right)
\end{array}
$$
\begin{aligned}
& \text { As, } 2 \sin ^2 \theta-5 \sin \theta+2>0 \\
& \Rightarrow(2 \sin \theta-1)(\sin \theta-2)>0 \\
& \quad[\text { where, }(\sin \theta-2) < 0 \text { for all } \theta \in R] \\
& \therefore \quad(2 \sin \theta-1) < 0
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & \sin \theta < \frac{1}{2}, \text { shown as } \\
\therefore & \theta \in\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, 2 \pi\right)
\end{array}
$$
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