Question 829571: 3-cosx=3sin^2x
cosx/1-sinx=1-sinx/cosx
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 3-cos(x) = 3sin^2(x)
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3 - cos(x) = 3(1-cos^2(x))
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3 - cos(x) = 3 - 3cos^2(x)
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3cos^2(x)-cos(x) = 0
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cos(x)[3cos(x)-1] = 0
cos(x) = 0 or cos(x) = 1/3
x = pi/2 or (3/2)pi or +-1.2310 radians
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cosx/1-sinx=1-sinx/cosx
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Cross-multiply to get:
cos^2(x) = 1-2sin(x)+sin^2(x)
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1-sin^2(x) = 1 - 2sin(x) + sin^2(x)
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2sin^2(x) -2sin(x) = 0
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sin(x)[2sin(x)-1] = 0
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sin(x) = 0 or sin(x) = 1/2
x = 0 or pi or pi/6 or (5/6)pi
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Cheers,
Stan H.
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