Posted by Katie on .
√5sec²x2secx=5
i have to solve this equation on the interval [0, 2π).

MATH 
Reiny,
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 2π1.0495 = 5.234
x = 2.686 or 3.597