SOLUTION: I need help with solving cos^2x - cos2x = 0 on the interval {0,2π). (cos^2x = cos(squared)x) This is what I have tried: (1) cos^2x - 2cos^2x - 1 = 0 (From the double-

Algebra ->  Trigonometry-basics -> SOLUTION: I need help with solving cos^2x - cos2x = 0 on the interval {0,2π). (cos^2x = cos(squared)x) This is what I have tried: (1) cos^2x - 2cos^2x - 1 = 0 (From the double-      Log On


   

Question 439333: I need help with solving cos^2x - cos2x = 0 on the interval {0,2π).
(cos^2x = cos(squared)x)
This is what I have tried:
(1)
cos^2x - 2cos^2x - 1 = 0 (From the double-angle identity for cos2x)
cos^2x - 2cos^2x = 1
-cos^2x = 1
Is there anything else that I could have done?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
I need help with solving
cos^2x - cos2x = 0 on the interval {0,2π).
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Note: cos(2x) = 2cos^2(x)-1
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cos^2(x)-[2cos^2(x)-1] = 0
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-cos^2(x)+1 = 0
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cos^2(x) - 1 = 0
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Factor:
(cos(x)-1)(cos(x)+1) = 0
cos(x) = 1 or cos(x) = -1
x = 0 or x = pi
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Cheers,
Stan H.
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