Trigonometric Equations Question 6
Question: The general value of $ \theta $ satisfying $ {{\sin }^{2}}\theta +\sin \theta =2 $ is
[AMU 1996, 99]
Options:
A) $ n\pi +{{(-1)}^{n}}\frac{\pi }{6} $
B) $ 2n\pi +\frac{\pi }{4} $
C) $ n\pi +{{(-1)}^{n}}\frac{\pi }{2} $
D) $ n\pi +{{(-1)}^{n}}\frac{\pi }{3} $
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Answer:
Correct Answer: B
Solution:
- $ {{\sin }^{2}}\theta +\sin \theta -2=0\Rightarrow (\sin \theta -1),(\sin \theta +2)=0 $
$ \Rightarrow $ $ \sin \theta \ne -2 $ ,
$ \therefore \sin \theta =1=\sin \pi /2 $
$ \Rightarrow $ $ \theta =n\pi +{{(-1)}^{n}}\frac{\pi }{2} $ .
BETA
