Trigonometric Equations Question 266
Question: The general solution of $ \sin x-3\sin 2x+\sin 3x= $ $ \cos x-3\cos 2x+\cos 3x $ is
[IIT 1989]
Options:
A) $ n\pi +\frac{\pi }{8} $
B) $ \frac{n\pi }{2}+\frac{\pi }{8} $
C) $ {{(-1)}^{n}}\frac{n\pi }{2}+\frac{\pi }{8} $
D) $ 2n\pi +{{\cos }^{-1}}\frac{3}{2} $
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Answer:
Correct Answer: B
Solution:
- $ \sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x $
$ \Rightarrow $ $ 2\sin 2x\cos x-3\sin 2x-2\cos 2x\cos x+3\cos 2x=0 $
$ \Rightarrow $ $ \sin 2x(2\cos x-3)-\cos 2x(2\cos x-3)=0 $
$ \Rightarrow $ $ (\sin 2x-\cos 2x)(2\cos x-3)=0\Rightarrow \sin 2x=\cos 2x $
$ \Rightarrow $ $ 2x=2n\pi \pm ( \frac{\pi }{2}-2x ) $ i.e., $ x=\frac{n\pi }{2}+\frac{\pi }{8} $ .
BETA
