SOLUTION: Solve the equation for x if 0 ≤ x < 2π. Give your answer in radians using exact values only. (Enter your answers as a comma-separated list.)
2 sin2 x + cos x − 1
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Question 1029573: Solve the equation for x if 0 ≤ x < 2π. Give your answer in radians using exact values only. (Enter your answers as a comma-separated list.)
2 sin2 x + cos x − 1 = 0
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Solve the equation for x if 0 ≤ x < 2π. Give your answer in radians using exact values only. (Enter your answers as a comma-separated list.)
2 sin^2(x) + cos x − 1 = 0
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2(1-cos^2(x)) + cos(x) -1 = 0
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-2cos^2(x) + cos(x) +1 = 0
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cos^2(x) - cos(x) - 1 = 0
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Use the quadratic formula to find potential values of cos(x)
cos(x) = [1 +- sqrt(1-4*-1)]/2 = [1+-sqrt(5)]/2
cos(x) = (1+sqrt(5))/2 :: extraneous
cos(x) = (1-sqrt(5))/2 = -0.6180
x = arccos(-0.6180) = 2.2370 radians or 4.0462 radians
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Cheers,
Stan H.
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