SOLUTION: solve over the interval [0,2pi) 2sin^2(x)+cos(x)-1=0

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Question 635251: solve over the interval [0,2pi)
2sin^2(x)+cos(x)-1=0
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
solve over the interval [0,2pi)
2sin^2(x)+cos(x)-1=0
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2(1-cos^2(x))+cos(x)-1 = 0
2 - 2cos^2 + cos - 1 = 0
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2cos^2 - cos -1 = 0
Factor:
(2cos+1)(cos-1) = 0
cos(x) = -1/2 or cos(x) = 1
x = (2/3)pi or x = (4/3)pi or x = 0
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Cheers,
Stan H.