Trigonometric Equations Question 328
Question: If $ {{\sin }^{2}}\theta =\frac{1}{4}, $ then the most general value of $ \theta $ is
[MNR 1973; MP PET 1984, 90]
Options:
A) $ 2n\pi \pm {{(-1)}^{n}}\frac{\pi }{6} $
B) $ \frac{n\pi }{2}\pm {{(-1)}^{n}}\frac{\pi }{6} $
C) $ n\pi \pm \frac{\pi }{6} $
D) $ 2n\pi \pm \frac{\pi }{6} $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ {{\sin }^{2}}\theta =\frac{1}{4}= $ $ {{\sin }^{2}}\frac{\pi }{6}\Rightarrow \theta =n\pi \pm \frac{\pi }{6} $ .
BETA
