Question

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

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

Answer 1

Answer: (D)

$(\frac{5\pi }{4},\frac{7\pi }{4})$

We have, ${\cos }^2\theta +\sin \theta +1=0$
$\Rightarrow 1-{\sin }^2\theta +\sin \theta +1=0$
$\Rightarrow {\sin }^2\theta -\sin \theta -2=0$
$\Rightarrow (\sin \theta +1)(\sin \theta -2)=0$
$\Rightarrow \sin \theta +1=0$
$\Rightarrow \sin \theta =-1$ $[\because \sin \theta \ne 2]$
$\Rightarrow \theta =\frac{3\pi }{2}\in (\frac{5\pi }{4},\frac{7\pi }{4})$