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

Answer 1

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