Use squaring to solve the following equation over the interval [0,2pi)

sin2x=1-cos2x

sin2x = 1-cos2x
sin²2x = 1 - 2cos2x + cos²2x
1 - cos²2x = 1 - 2cos2x+ cos²2x
0 = -2cos2x + 2cos²2x
cos2x(cos2x-1) = 0
cos2x = 0 ==> 2x = π/2 + nπ = (1+2n)π/2 ==> x = (1+2n)π/4
cos2x = 1 ==> 2x = 2nπ ==> x = nπ

x = (1+2n)π/4 = π/4, 3π/4, 5π/4, 7π/4
x= nπ = 0, π

x = 0, π/4, 3π/4, π, 5π/4, 7π/4
But, squaring the equation introduces the possibility of extraneous solutions, so the solutions must be verified.
x = 0 : sin0 = 1-cos0 : 0 = 0 : true solution
x = π/4 : sinπ/2 = 1-cosπ/2 : 1 = 1 : true solution
x = 3π/4 : sin3π/2 = 1 - cos3π/2 : -1 = 1 : extraneous solution
x = π : sin2π = 1 - cos2π : 0 = 0 : true solution
x = 5π/4 : sin5π/2 = 1 - cos5π/2 : 1 = 1 : true solution
x = 7π/4 : sin7π/2 = 1 - cos7π/2 : -1 = 1 : extraneous solution

After purging the extraneous solutions, the solution set on the interval [0, 2π) becomes:
x = 0, π/4, π, 5π/4

I think MechEn's approach better than squaring; the only item amiss in Mech...'s solution is that 2π was included though your interval did not.

Squaring... [sin2x +cos2x]² = 1² gives {sin²(2x) + cos²(2x)} + 2sin2xcos2x = 1
The quantity in braces { } is one so you end up with

2(sin2x)cos2x = 0 which is sin(4x) = 0 or 4x = 0 + nπ

which generates 8 possible x's, 4 of which fail in the original equation, leaving you with this solution set: {0, π/4, π, 5π/4 } from the finite closed-open interval [0, 2π) you restricted your answer to.

2*sin(x)*cos(x) = 1 - cos²(x) + sin²(x)

2*sin(x)*cos(x) = 2*sin²(x)

2*sin²(x) - 2*sin(x)*cos(x) = 0

2*sin(x)(sin(x) - cos(x)) = 0

sin(x) = 0 or tan(x) = 1

x = 0, pi, 2pi, pi/4, 5pi/4

It's unclear whether you mean sin(2x) or sin^2x.
If you mean sin^2x = 1 - cos^2x, then this is true for all x.
The solution would be all x in [0, 2pi).

sin2x+cos2x = 1 ---> sin(2x + (pi/4)) = 1/sqrt2 , then easy to solve for x