Solve the equation sin2x-sin4x=0?

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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}

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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}

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