`csc^2 (x) + 3csc(x) - 4 = 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 73 - Precalculus (3rd Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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This is a quadratic equation for csc(x). The roots are 1 and -4.

csc(x)=1/sin(x), so we have two possibilities:

sin(x)=1 or sin(x)=-1/4.

The roots of these equations in `(0, 2pi)` are

`pi/2,` `pi+arcsin(1/4)` and `2pi-arcsin(1/4).`

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