but its not satisfying any of the options
4 Answers
Asish Mahapatra
·2010-02-21 04:45:43
sin6x + sin2x = sin4x
=> 2sin4xcos2x - sin4x=0
=> sin4x(2cos2x-1) = 0
=>4x=n\pi
OR
2x=2n\pi \pm \pi /3
nihal raj
·2010-11-09 22:07:48
sin6x = 2 sin(4x-2x)/2 . cos (4x+2x)/2
sin2(3x) = 2 sinx.cos3x
2 sin3x.cos3x= 2 sinx.cos3x
sin 3x = sinx
3sinx - 4 sin3x = sinx
3- 4sin2x=1
sin2x=1/2
sin x = + - 1/√2
looking at the option b satisfies above conditions...at n = + - 3
hence i think it should be option b)