Question 248490: Solve for 'x'
0=Cos^2(x)-2Sinx
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Solve for 'x'
0=Cos^2(x)-2Sinx
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cos^2(x) = 1 - sin^2(x)
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Equation:
0 = 1-sin^2(x) - 2sin(x)
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= 0Rearrange:
sin^2(x) + 2sin(x) -1 = 0
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This is a quadratic with a=1; b = 2 ; c = -1
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sin(x) = [-2 +- sqrt(4 - 4*1*-1)]/2
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sin(x) = [-2 +- sqrt(8)]/2
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sin(x) = [-2 +- 2sqrt(2)]/2
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sin(x) = [-1 + sqrt(2)] or sin(x) = [-1-sqrt(2)]
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Since -1<= sin(x) <= 1, only sin(x) = -1+sqrt(2) is a solution:
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sin(x) = -1 + 1.414 = 0.4142
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x = sin^-1(0.4142) = 24.47 degrees
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Cheers,
Stan H.
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