Trigonometric Equations Question 34
Question: If $ \sin 5x+\sin 3x+\sin x=0 $ , then the value of x other than 0 lying between $ 0\le x\le \frac{\pi }{2} $ is
[MNR 1985]
Options:
A) $ \frac{\pi }{6} $
B) $ \frac{\pi }{12} $
C) $ \frac{\pi }{3} $
D) $ \frac{\pi }{4} $
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Answer:
Correct Answer: C
Solution:
- $ \sin 5x+\sin 3x+\sin x=0 $
$ \Rightarrow $ $ -\sin 3x=\sin 5x+\sin x=-2\sin 3x\cos 2x $ $ \Rightarrow $ $ \sin 3x=0 $
$ \Rightarrow $ $ x=0 $ or $ \cos 2x=-\frac{1}{2}=-\cos ,( \frac{\pi }{3} )=\cos ,( \pi -\frac{\pi }{3} ) $
$ \Rightarrow $ $ 2x=2n\pi \pm ( \pi -\frac{\pi }{3} ),\Rightarrow x=n\pi \pm ( \frac{\pi }{3} ) $ For x lying between 0 and $ \frac{\pi }{2} $ , we get $ x=\frac{\pi }{3} $ and $ x=0 $ . Trick: Check with options.
BETA
