Trigonometric Equations Question 67
Question: The solution of equation $ {{\cos }^{2}}\theta +\sin \theta +1=0 $ lies in the interval
[UPSEAT 2004; IIT 1992]
Options:
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} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
- We have, $ {{\cos }^{2}}\theta +\sin \theta +1=0 $
Þ $ 1-{{\sin }^{2}}\theta +\sin \theta +1=0 $
Þ $ {{\sin }^{2}}\theta -\sin \theta -2=0 $
Þ $ (\sin \theta +1),(\sin \theta -2)=0 $ $ \sin \theta =2 $ , which is not possible and $ \sin \theta =-1 $ . Therefore, solution of given equation lies in the interval $ ( \frac{5\pi }{4},,\frac{7\pi }{4} ) $ .
BETA
