find the soln..........

sin6x + sin2x = sin4x
=> 2sin4xcos2x - sin4x=0
=> sin4x(2cos2x-1) = 0

=>4x=n\pi

OR

2x=2n\pi \pm \pi /3

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)

b is correct