Trigonometric Equations Question 382
Question: The general value of $ \theta $ satisfying the equation $ 2{{\sin }^{2}}\theta -3\sin \theta -2=0 $ is
[Roorkee 1993]
Options:
A) $ n\pi +{{(-1)}^{n}}\frac{\pi }{6} $
B) $ n\pi +{{(-1)}^{n}}\frac{\pi }{2} $
C) $ n\pi +{{(-1)}^{n}}\frac{5\pi }{6} $
D) $ n\pi +{{(-1)}^{n}}\frac{7\pi }{6} $
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Answer:
Correct Answer: D
Solution:
- $ 2{{\sin }^{2}}\theta -3\sin \theta -2=0 $
$ \Rightarrow $ $ (2\sin \theta +1)(\sin \theta -2)=0 $
$ \Rightarrow $ $ \sin \theta =-\frac{1}{2} $ , ( $ \because ,\sin \theta \ne 2) $
$ \Rightarrow $ $ \sin \theta =\sin ( \frac{-\pi }{6} ) $
$ \Rightarrow $ $ \theta =n\pi +{{(-1)}^{n}}( \frac{-\pi }{6} )\Rightarrow \theta =n\pi +{{(-1)}^{n+1}}\frac{\pi }{6} $
$ \Rightarrow $ $ \theta =n\pi +{{(-1)}^{n}}\frac{7\pi }{6} $ , $ { \because \frac{-\pi }{6}\text{is equivalent to }\frac{7\pi }{6} } $ .
BETA
