Question

The solution of the equation $\cos ^{2} \theta+\sin \,\theta+1=0$, lies in the interval

• A. $\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$
• B. $\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$
• C. $\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right)$
• D. $\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right)$

Answer 1

Answer: (D)

We have $\sin ^{2} \theta-\sin\, \theta-2=0$
$ \Rightarrow (\sin \,\theta+1)(\sin\, \theta-2)=0$
As $\sin\, \theta \neq 2$
$ \therefore \sin\, \theta=-1=\sin \frac{3 \pi}{2}$
$\therefore \theta=\frac{3 \pi}{2}=\frac{6 \pi}{4} \in\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right)$