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August 28, 2014

August 28, 2014

Posted by **Katie** on Wednesday, February 22, 2012 at 9:16pm.

i have to solve this equation on the interval [0, 2π).

- MATH -
**Reiny**, Wednesday, February 22, 2012 at 10:40pmlet 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 2π-1.0495 =**5.234**

x =**2.686**or**3.597**

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