Solutions in the interval for sin^2x-cos^2x=0

 cos^2 x - sin^2 x = 0 
cos 2x = 0 
x = 45 and 135 
Add or subtract multiples of 360. Those are valid x values also. 

Sin²x -cos²x=0

Remember:
sin²x + cos²x =1    ⇒ sin²x=1-cos²x

Therefore:
sin²x -cos²x=0
(1-cos²x)-cos²x=0
-2cos²x=-1
cos² x=-1/(-2)
cos²x=1/2
cos x=⁺₋√(1/2)
cos x=⁺₋(√2)/2

We have two solutions:
Solution 1:
x=cos⁻¹ -(√2)/2=3π/4  + Kπ     or  135º + 180ºk

Solution 2:
x=cos⁻¹ (√2)/2=π/4 +Kπ           or       45º+180ºk         (k=...-2,-1,0,1,2...)

Solution=solution 1  U  solution 2=π/4+π/2 K   or  45º+90ºK

π/4+π/2 K   or  45º+90ºK

The correct answer is $90