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The solution of the equation $\cos ^{2} \theta+\sin \theta+1=0$, lies in the interval
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2204 Upvotes
Verified Answer
The correct answer is:
$\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right)$
We have,
$$
\begin{array}{l}
\cos ^{2} \theta+\sin \theta+1=0 \Rightarrow 1-\sin ^{2} \theta+\sin \theta+1 \\
=0 \\
\Rightarrow \quad \sin \theta=-1(\because \sin \theta \neq 2) \Rightarrow \theta=3 \pi / 2 \\
\therefore \theta \in\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right)
\end{array}
$$
$$
\begin{array}{l}
\cos ^{2} \theta+\sin \theta+1=0 \Rightarrow 1-\sin ^{2} \theta+\sin \theta+1 \\
=0 \\
\Rightarrow \quad \sin \theta=-1(\because \sin \theta \neq 2) \Rightarrow \theta=3 \pi / 2 \\
\therefore \theta \in\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right)
\end{array}
$$
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