Question

The solution of 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)

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

$co{s}^2\theta +\sin \theta +1=0$

$\Rightarrow 1-{\sin }^2\theta +\sin \theta +1=0$

$\Rightarrow {\sin }^2\theta -\sin \theta -2=0$

$\Rightarrow (\sin \theta +1)(\sin \theta -2)=0$

As $-1<\sin \theta <1$ so $\sin \theta +1=0$

$\Rightarrow \sin \theta =\sin \frac{3\pi }{2}$

$\Rightarrow \theta =\frac{3\pi }{2}\epsilon (\frac{5\pi }{4},\frac{7\pi }{4})$