Solve for x : sin4x-sin2x=0

To solve for x . sin4x -sin2x =
0.

To solve the equation we use sin2A =
2sinAcosA.

Therefore ,  sin4x = sin2(2x) =
2sin2x*cos2x.

So we substitute sin4x = 2sin2x cos2x 9n the
given equation:

2sin2x*cos2x - sin2x =
0

sin2x*(2cos2x -1) =
0..............(1).

Again  we  substitute cos2x =
1-(2sinx)^2 and sin2x = 2sinxcosx on (1):

2sinx*cosx
{2(1-2(sinx)^2) -1} = 0

2 sinx*cosx{1 - 4(sinx)^2} =
0

Equate each  non numeriacal factors to
zero:

sinx = 0, cosx = 0 , 1-4(sinx)^2 =
0

So sin x = 0 gives: x= 0 or
pi

cosx = 0 gives x = pi/2 or
3pi/2

1-4(sinx)^2 = gives: sinx = 1 or =-1. So x = pi/4,
3pi/4 , 5pi/4 or 7pi/4