Trigonometric Identities Question 208
Question: The number of solutions of $ \sin x+\sin 2x+\sin 3x=\cos x+\cos 2x+\cos 3x, $ $ 0\le x\le 2\pi $ , is
Options:
A) 7
B) 5
C) 4
D) 6
Show Answer
Answer:
Correct Answer: D
Solution:
[d] we have $ (sinx+sin3x)+sin2x=(cosx+cos3x)+cos2x $ Or $ 2\sin 2x\cos x+\sin 2x=2\cos 2x\cos x+\cos 2x $ Or $ \sin 2x(2cosx+1)=cos2x(2cosx+1) $ Or $ (2cosx+1)(sin2x-\cos 2x)=0 $ Or $ \cos x=-1/2\sin 2x-\cos 2x=0 $
$ \Rightarrow x=2n\pi \pm (2\pi /3)or,tan2x=1=tan(\pi /4) $ $ =2n\pi \pm (2\pi /3)or,x=(4n+1)\pi /8,n\in Z $ But here $ 0\le x\le 2\pi . $ Hence, $ x=\pi /8,5\pi /8,2\pi /3,9\pi /8,4\pi /3,13\pi /8. $
BETA
