Trigonometric Equations Question 3
Question: $ \sin 6\theta +\sin 4\theta +\sin 2\theta =0, $ then $ \theta = $
[MP PET 1999; Pb. CET 2000]
Options:
A) $ \frac{n\pi }{4} $ or $ n\pi \pm \frac{\pi }{3} $
B) $ \frac{n\pi }{4} $ or $ n\pi \pm \frac{\pi }{6} $
C) $ \frac{n\pi }{4} $ or $ 2n\pi \pm \frac{\pi }{6} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \sin 6\theta +\sin 4\theta +\sin 2\theta =0 $
$ \Rightarrow 2\sin 4\theta \cos 2\theta +\sin 4\theta =0 $
$ \Rightarrow $ $ \sin 4\theta (2\cos 2\theta +1)=0 $
$ \Rightarrow $ $ 2\cos 2\theta =-1 $
$ \Rightarrow $ $ \cos 2\theta =-\frac{1}{2} $
$ \Rightarrow $ $ 2\theta =2n\pi \pm \frac{2\pi }{3}\Rightarrow \theta =n\pi \pm \frac{\pi }{3} $ and $ \sin 4\theta =0\Rightarrow 4\theta =n\pi \Rightarrow \theta =\frac{n\pi }{4} $ $ \theta =\frac{n\pi }{4} $ or $ n\pi \pm \frac{\pi }{3} $ .
BETA
