Trigonometric Identities Question 273
Question: $ {{| \cos x |}^{{{\sin }^{2}}x-\frac{3}{2}\sin x+\frac{1}{2}}}=1, $ then possible values of x :
Options:
A) $ n,\pi $ or $ 2n,\pi +\frac{\pi }{2} $
B) $ n,\pi $ or $ 2n\pi +\frac{\pi }{2} $ or $ n\pi +{{(-1)}^{n}}\frac{\pi }{6}, $ $ n\in I $
C) $ n\pi +{{(-1)}^{n}}\frac{\pi }{6},,n\in I $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
The equation holds if $ |\cos x|=1 $ i.e., if $ x=n\pi ,n\in I $ If $ |\cos x|\ne 1 $ then $ {{\sin }^{2}}x-\frac{3}{2}\sin x+\frac{1}{2}=0 $
$ \Rightarrow \sin x=1 $ or $ \frac{1}{2} $ $ \sin x\ne 1, $ as in that case $ cosx=0 $
$ \therefore \sin x=\frac{1}{2}\Rightarrow x=n\pi +{{(-1)}^{n}}\frac{\pi }{6} $
BETA
