A) 7
B) 5
C) 4
D) 6
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, \[x=\pi /8,\,\,5\pi /8,\,\,2\pi /3,\,\,9\pi /8,\,\,4\pi /3,\,\,13\pi /8.\]You need to login to perform this action.
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