Trigonometric Equations Question 50
Question: If $ 2{{\sin }^{2}}\theta =3\cos \theta , $ where $ 0\le \theta \le 2\pi $ , then $ \theta = $
[IIT 1963]
Options:
A) $ \frac{\pi }{6},\frac{7\pi }{6} $
B) $ \frac{\pi }{3},\frac{5\pi }{3} $
C) $ \frac{\pi }{3},\frac{7\pi }{3} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- $ 2-2{{\cos }^{2}}\theta =3\cos \theta $
Þ $ 2{{\cos }^{2}}+3\cos \theta -2=0 $
Þ $ \cos \theta =\frac{-3\pm \sqrt{9+16}}{4}=\frac{-3\pm 5}{4} $ Neglecting (-) sign, we get $ \cos \theta =\frac{1}{2}=\cos ( \frac{\pi }{3} ) $
$ \Rightarrow $ $ \theta =2n\pi \pm \frac{\pi }{3} $ . The values of $ \theta $ between 0 and $ 2\pi $ are $ \frac{\pi }{3},\frac{5\pi }{3} $ .
BETA
