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

Answer: π/4+π/2 K or 45º+90ºK