Posted by **Courtney** 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 -
**Reiny**, 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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