`2sin^2 (x) + 5cos(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 67 - Precalculus (3rd Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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First transform `sin^2(x)` into `1-cos^2(x)` and obtain

`2-2cos^2(x)+5cos(x)=4,`

`2cos^2(x)-5cos(x)+2=0.`

This is a quadratic equation for cos(x), the roots are 2 and 1/2. cos(x) never =2.

cos(x)=1/2 on `(0, 2pi)` at `x_1=pi/3` and `x_2=(5pi)/3.`

The answer: `pi/3` and `(5pi)/3.`

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