Trigonometric Equations Question 23
Question: The roots of the equation $ 1-\cos \theta =\sin \theta .\sin \frac{\theta }{2} $ is
[Orissa JEE 2004]
Options:
A) $ k\pi ,k\in I $
B) $ 2k\pi ,k\in \mathbb{Z} $
C) $ k\frac{\pi }{2},k\in \mathbb{Z} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- We have, $ 1-\cos \theta =2\sin^2 \frac{\theta }{2} $
Þ $ 2{{\sin }^{2}}\frac{\theta }{2}=2\sin \frac{\theta }{2},.,\cos \frac{\theta }{2},.,\sin \frac{\theta }{2} $
Þ $ 2{{\sin }^{2}}\frac{\theta }{2},[ 1+\cos \frac{\theta }{2} ]=0 $ Þ $ \sin \frac{\theta }{2}=0 $ or $ 2{{\sin }^{2}}\frac{\theta }{4}=0 $
Þ $ \sin \frac{\theta }{2} = 0 $ or $ \sin \frac{\theta }{4} = 0 $ Þ $ \frac{\theta }{2}=k\pi $ or $ \frac{\theta }{4}=k\pi $ . Hence, $ \theta =2k\pi $ or $ \theta =4k\pi $ , $ k\in \mathbb{Z} $ .
BETA
