Trigonometric Equations Question 61
Question: If $ (2\cos x-1)(3+2\cos x)=0,,0\le x\le 2\pi $ , then $ x= $
[MNR 1988; UPSEAT 2000]
Options:
A) $ \frac{\pi }{3} $
B) $ \frac{\pi }{3},\frac{5\pi }{3} $
C) $ \frac{\pi }{2},\frac{5\pi }{3},{{\cos }^{-1}}( -\frac{3}{2} ) $
D) $ \frac{5\pi }{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ (2\cos x-1)(3+2\cos x)=0 $ Then $ \cos x=\frac{1}{2}\text{ as},\cos x\ne \frac{-3}{2} $
$ \Rightarrow $ $ x=2n\pi \pm \frac{\pi }{3};{ \begin{aligned} & \text{ for }n=0,x=\frac{\pi }{3},,\frac{5\pi }{3} \\ & \text{ for }n=1,x=\frac{5\pi }{3} \\ \end{aligned} } $
BETA
