Trigonometry factoring how to

(cos(2x))^2-(sin(2x))^2+sin(2x)=0
I'm not even sure where to start. I tried factoring our sin2x, but then I couldn't figure out what to do with cos^2 -sin2x

sin^2 + cos^2 = 1 or cos^2 = 1 - sin^2. Substituting this expression into the equation,

cos^2 - sin^2 + sin = 0 is equivalent to
(1 - sin^2) - sin^2 + sin = 0 iff
-2sin^2 + sin + 1 = 0 iff
2sin^2 - sin - 1 = 0 iff
(2sin + 1)(sin - 1) = 0 iff
2sin + 1 = 0 or sin - 1 = 0 iff
sin(2x) = -1/2 or sin(2x) = 1

Solve from here.

(cos(2x))^2-(sin(2x))^2+sin(2x)=0
cos(4x) + sin(2x) = 0 { because cos(2x) = cos^2(x) - sin^2(x)}
sin(pi/2 -4x) + sin(2x) =0 { because sin(90-theta) = cos(theta)
Apply product rule
2 sin{(pi/2 -4x+2x)/2}* cos{(pi/2 -4x-2x)/2}=0
sin(pi/4- x) *cos(pi/4 -3x) = 0....................Ans