How do i solve sin(x)^2-cos(x)^2=0 ,, 0 ≤ x < 2π

(sin x) ^2 - (cos x)^2= 0
( sin x - cos x) ( sin x + cos x) =0
sin x - cos x = 0
sin x = cos x
x = pi/4 and 5pi/4

sin x + cos x = 0
sin x = - cos x
x = 3pi/4 and 7pi/4

Well, try this:

1. You should know by this stage that sin (x) ^2 + cos (x) ^2 = 1

2. Solve for sin (x) ^ 2 to give sin (x) ^2 = 1 - cos (x) ^2

3. Substitute 2 back into 1:
( 1 - cos (x) ^2 ) - cos (x) ^2 = 0
1 - 2 cos (x) ^2 = 0
cos (x) ^2 = 1/2
cos (x) ^2 = 1/2
Square root both sides:
cos (x) = (1 / sqrt(2))

Now, I have done the tough part, you can work out the rest. All you need to do is find out for what values of x, in between 0 and 2 pi, does cos (x) = 1 / (sqrt(2))

Please invest the same amount of time I have to finish this yourself.

Use half-angle identity

1/2(1-cos2x)-1/2(1-cos2x)=0
Times by 2 and collect

1-1+2cos2x=0
Divide by 2, and inverse

(cos0)^-1=2x ((cos0)^-1)/2=x
Finally, arccos at 0 equals pi/2, simplify, and

X=pi/4
Test (also see if fits limit, 0
Sin(pi/4)^2-cos(pi/4)^2=0 Yes