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
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sin^2 - (1-sin^2) = 0
2sin^2 - 1 = 0
sin = ħsqrt(1/2)
x = pi/4, 3pi/4, 5pi/4, 7pi/4
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Or,
sin^2x-cos^2x=0
(sin + cos)*(sin - cos) = 0
sin = cos
sin = -cos
Same as above
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