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If $\sqrt{2} \sin ^2 x+(3 \sqrt{2}+1) \sin x+3>0$ and $x^2-7 x+10 < 0$, then $x$ lies in the interval
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Verified Answer
The correct answer is:
$\left(2, \frac{5 \pi}{4}\right)$
We have,
$$
\begin{array}{ll}
& \sqrt{2} \sin ^2 x+(3 \sqrt{2}+1) \sin x+3>0 \\
\Rightarrow & (\sqrt{2} \sin x+1)(\sin x+3)>0 \\
\Rightarrow & \sin x>\frac{-1}{\sqrt{2}} x \in\left(\pi, \frac{5 \pi}{4}\right) \\
\text { and } & x^2-7 x+10 < 0 \\
& (x-5)(x-2) < 0 \\
& x \in(2,5)
\end{array}
$$
and
From Eqs. (i) and (ii), we get
$$
x \in\left(2, \frac{5 \pi}{4}\right)
$$
$$
\begin{array}{ll}
& \sqrt{2} \sin ^2 x+(3 \sqrt{2}+1) \sin x+3>0 \\
\Rightarrow & (\sqrt{2} \sin x+1)(\sin x+3)>0 \\
\Rightarrow & \sin x>\frac{-1}{\sqrt{2}} x \in\left(\pi, \frac{5 \pi}{4}\right) \\
\text { and } & x^2-7 x+10 < 0 \\
& (x-5)(x-2) < 0 \\
& x \in(2,5)
\end{array}
$$
and
From Eqs. (i) and (ii), we get
$$
x \in\left(2, \frac{5 \pi}{4}\right)
$$
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