`2cos^2 (x) + 7sin(x) = 5` Use inverse functions where needed to find all solutions of the equation in the interval `0,2pi)`.

Textbook Question

Chapter 5, 5.3 - Problem 68 - Precalculus (3rd Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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

`2-2sin^2(x)+7sin(x)=5,`

`2sin^2(x)-7sin(x)+3=0.`

This is a quadratic equation for sin(x), the roots are

sin(x)=3 and sin(x)=1/2.

The first is impossible, the second has two roots on `(0, 2pi)`, `pi/6` and `(5pi)/6.`

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