`sec^2 (x) + 2sec(x) - 8 = 0` Use inverse functions where needed to find all solutions of the equation in the interval `0,2pi)`.

Textbook Question

Chapter 5, 5.3 - Problem 72 - Precalculus (3rd Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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Denote sec(x) as y and obtain a quadratic equation for y:

`y^2+2y-8=0.`

This equation has two roots, `y_1=2` and `y_2=-4.` So we have to solve

sec(x)=2 and sec(x)=-4 separately.

sec(x) = 1/cos(x), so cos(x)=1/2 or cos(x)=-1/4.

Each of these equations has two roots on `(0, 2pi):`

`x_1 = pi/3,` `x_2 = (5pi)/3,` `x_3 = cos^(-1)(-1/4),` `x_4 = 2pi -cos^(-1)(-1/4).`

(`x_1=60°,` `x_2=300°,` `x_3 approx 104.5°,` `x_4 approx 255.5°` degrees)

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