`sec^2 (x) - sec(x) = 2` Find all the solutions of the equation in the interval `[0,2pi)`.

Textbook Question

Chapter 5, 5.3 - Problem 29 - Precalculus (3rd Edition, Ron Larson).
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mathace | (Level 3) Assistant Educator

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Solve the equation `sec^2(x)-sec(x)=2,[0,2pi).`

`sec^2(x)-sec(x)-2=0`

`(sec(x)-2)(sec(x)+1)=0` 

Set each factor equal to zero and solve for the x values.

`sec(x)-2=0`

`sec(x)=2`

`cos(x)=1/2`

`x=pi/3,x=(5pi)/3`

`sec(x)+1=0`

`sec(x)=-1`

`cos(x)=-1`

`x=pi`

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