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
