Trigonometric Identities Question 211
Question: The number of solution of equation $ 6\cos 2\theta +2{{\cos }^{2}}(\theta /2)+2sin^{2}\theta =0, $ $ -\pi <\theta <\pi $ is
Options:
A) 3
B) 4
C) 5
D) 6
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ 6\cos 2\theta +2{{\cos }^{2}}(\theta /2)2sin^{2}\theta =0 $ Or $ 12{{\cos }^{2}}\theta -6+1+\cos \theta +2-2{{\cos }^{2}}\theta =0 $ or $ 10{{\cos }^{2}}\theta +\cos gq-3=0 $ or $ (5cos\theta +3)(2cos\theta -1)=0 $ or $ \cos \theta =-\frac{3}{5},\frac{1}{2} $
$ \Rightarrow \theta =\frac{\pi }{3};\pi -{{\cos }^{-1}}( \frac{3}{5} ),-\frac{\pi }{3},-\pi +{{\cos }^{-1}}( \frac{3}{5} ) $
BETA
