Trigonometric Equations Question 58
Question: If $ 5\cos 2\theta +2{{\cos }^{2}}\frac{\theta }{2}+1=0,-\pi <\theta <\pi $ , then $ \theta = $
[Roorkee 1984]
Options:
A) $ \frac{\pi }{3} $
B) $ \frac{\pi }{3},{{\cos }^{-1}}\frac{3}{5} $
C) $ {{\cos }^{-1}}\frac{3}{5} $
D) $ \frac{\pi }{3},\pi -{{\cos }^{-1}}\frac{3}{5} $
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Answer:
Correct Answer: D
Solution:
- $ 5\cos 2\theta +2{{\cos }^{2}}\frac{\theta }{2}+1=0 $
$ \Rightarrow $ $ 5(2{{\cos }^{2}}\theta -1)+(1+\cos \theta )+1=0 $
$ \Rightarrow $ $ 10{{\cos }^{2}}\theta +\cos \theta -3=0 $
$ \Rightarrow $ $ (5\cos \theta +3),(2\cos \theta -1)=0 $
$ \Rightarrow $ $ \cos \theta =\frac{1}{2},,\cos \theta =-\frac{3}{5}\Rightarrow \theta =\frac{\pi }{3},,\pi -{{\cos }^{-1}}( \frac{3}{5} ) $ .
BETA
