# 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
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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