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Posted by on Wednesday, May 12, 2010 at 12:11pm.

Given 2secant^2 (theta) + 1 = 5 secant (theta), solve for (theta), to the nearest degree, in the interval 0 degrees is greater than or equal to (theta) which is less than 360 degrees

  • Trig - , Wednesday, May 12, 2010 at 1:20pm

    let sec(theta) = x
    so we have
    2secant^2 (theta) + 1 = 5 secant (theta) becomes
    2x^2 + 1 = 5x
    2x^2 - 5x + 1 = 0
    x = (5 ± √17)/4
    = 2.2808 or .21922
    so sec(theta) = 2.2808
    cos(theta) = 1/2.2808 = .43844 , so theta is in I or IV
    theta = 63.995 or 64º or theta = 296º

    or

    sec(theta) = .21922
    cos(theta) = 1/.21922 = 4.56 ---> not possible
    theta = 64º or 296º

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