`sec^2 (x) - 6tan(x) = -4` Use inverse functions where needed to find all solutions of the equation in the interval `0,2pi)`.

Textbook Question

Chapter 5, 5.3 - Problem 65 - Precalculus (3rd Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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First, transform `sec^2(x)` into `1+tan^(x)` (this is obvious):

`1+tan^2(x)-6tan(x)=-4,` or `tan^2(x)-6tan(x)+5=0.`

This is a quadratic equation for tan(x), its roots are 1 and 5.

So we have tan(x)=1 or tan(x)=5.

On `(0, 2pi)` there are four roots: `pi/4,` `(5pi)/4,` `arctan(5)` and `pi+arctan(5).`

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