MATH

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√5sec²x-2secx=5

i have to solve this equation on the interval [0, 2π).

  • MATH -

    let secx = y , so we have

    √5y^2 - 2y - 5 = 0
    using the quad formula,
    a = √5 , b = -2, and c = -5
    y = (2 ± √(4 - 4(√5)(-5) )/2√5
    = (2 ± √(4+20√5) )/(2√5)
    = appr 2.008 or -1.114

    then since secx = 1/cosx
    cosx = .498 or cosx = -.898
    setting my calculator to radians ...
    x = 1.0495 or 2π-1.0495 = 5.234
    x = 2.686 or 3.597

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