SOLUTION: Solve trig equation on the interval [0,2pi]: sin^2x-cos^2x=0

Question 1036995: Solve trig equation on the interval [0,2pi]:
sin^2x-cos^2x=0
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Solve trig equation on the interval [0,2pi]:
sin^2x-cos^2x=0
---
sin^2 - (1-sin^2) = 0
2sin^2 - 1 = 0
sin = �sqrt(1/2)
x = pi/4, 3pi/4, 5pi/4, 7pi/4
=======================
Or,
sin^2x-cos^2x=0
(sin + cos)*(sin - cos) = 0
sin = cos
sin = -cos
Same as above

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