Trigonometric Identities Question 214
Question: Let $ 0<{\theta_1}<{\theta_2}<{\theta_3}<… $ denote the positive solution of the equation $ 3+3\cos \theta =2{{\sin }^{2}}\theta $ The value of $ {\theta_3}+{\theta_7} $ is
Options:
A) 6 $ \pi $
B) 7 $ \pi $
C) 8 $ \pi $
D) 4 $ \pi $
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Answer:
Correct Answer: A
Solution:
[a] $ 3+3\cos \theta =2-2{{\cos }^{2}}\theta $
$ \Rightarrow 2{{\cos }^{2}}\theta +3\cos \theta +1=0 $
$ \Rightarrow (2cos\theta +1)(cos\theta +1)=0 $
$ \therefore \cos \theta =-1or\cos \theta =-\frac{1}{2} $ if $ \cos \theta =-1 $ then $ \theta =\pi ,3\pi ,5\pi ,7\pi ,9\pi ,… $ if $ \cos \theta =-1/2 $ then $ \theta =\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3},\frac{14\pi }{3},… $ Solutions in increasing order are $ 0<\frac{2\pi }{3}<\pi <\frac{4\pi }{3}<\frac{8\pi }{3}<3\pi <\frac{10\pi }{3}<\frac{14\pi }{3}<5\pi ,.. $
$ \therefore ,{\theta_3}+{\theta_7}=\frac{4\pi }{3}+\frac{14\pi }{3}=6\pi $
BETA
