SOLUTION: 1.Solve the equation 2cos x-1=0, 0<x<2pi 2. Solve the equation (cos^2)x-(sin^2)x=cos x, 0<x<2pi

Algebra ->  Trigonometry-basics -> SOLUTION: 1.Solve the equation 2cos x-1=0, 0<x<2pi 2. Solve the equation (cos^2)x-(sin^2)x=cos x, 0<x<2pi      Log On


   

Question 864465: 1.Solve the equation 2cos x-1=0, 0 2. Solve the equation (cos^2)x-(sin^2)x=cos x, 0
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1.Solve the equation
2cos x-1=0
cos(x) = 1/2
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x = pi/3 or x = -pi/3
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2. Solve the equation (cos^2)x-(sin^2)x=cos x, 0
2cos^2(x)-1 = cos(x)
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2cos^2(x) - cos(x) -1 = 0
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cos(x) = [-1 +- sqrt(1-4*2*-1)]/4
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cos(x) = [-1 +- 3]/4
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cos(x) = -3/4 or cos(x) = 1/2
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x = arccos(-3/4) or cos(x) = 1/2
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x = 138.59 or x = 60 degrees
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Cheers,
Stan H.
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