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

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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.
User Tarn
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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

User Blackbam
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