sin2x-sin4x=0

sinx-siny = 2cos(x+y)/2*sin(x-y)/2

==> sin2x-sin4x= 2cos(3x)*sin(-x) = 2cos3x*-sinx= -2cos3xsinx=0

==> -2cos3x*sinx= 0

then cos3x=0 OR sinx=0

cos3x=0

==> 3x= pi/2 + 2n*pi

==> x= pi/6 +(2*pi/3)n (n is integer)

sinx= 0

==> x= n*pi

==> x = {n*pi, pi/6+(2pi/3)n}

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

sin2x-sin4x=0

sinx-siny = 2cos(x+y)/2*sin(x-y)/2

==> sin2x-sin4x= 2cos(3x)*sin(-x) = 2cos3x*-sinx= -2cos3xsinx=0

==> -2cos3x*sinx= 0

then cos3x=0 OR sinx=0

cos3x=0

==> 3x= pi/2 + 2n*pi

==> x= pi/6 +(2*pi/3)n (n is integer)

sinx= 0

==> x= n*pi

==> x = {n*pi, pi/6+(2pi/3)n}