Question 853174: how many values of x are in the interval [0,2pi] for the equation 2cosx=tanx
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! how many values of x are in the interval [0,2pi] for the equation
2cosx=tanx
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2cos(x) = sin(x)/cos(x)
2cos^2(x) = sin(x)
2(1-sin^2(x)) = sin(x)
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2sin^2(x) + sin(x) -2 = 0
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sin(x) = [-1 +- sqrt(1 - 4*2*-2)]/4
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sin(x) = [-1 +- sqrt(17)]/4
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sin(x) = (-1-sqrt(17))/4 = -1/28 (extraneous since -1<= sin(x) <=))
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sin(x) = (-1+sqrt(17))/4 = 0.7808
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x = arcsin(0.7808) = 0.8959 radians
OR
x = arcsin(0.7808) = (pi-0.8959) = 2.2456 radians
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Cheers,
Stan H.
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