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

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Chapter 5, 5.3 - Problem 74 - Precalculus (3rd Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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`csc^2 (x) - 5csc(x) = 0`  is equivalent to `csc(x)*(csc(x)-5) = 0,` so

`csc(x)=0 or csc(x)=5.`

By definition, `csc(x)=1/sin(x)` and this is never zero.

So `sin(x) = 1/5` remains.

There are two solutions on `(0, 2pi),`

`x_1=arcsin(1/5)` and `x_2=pi-arcsin(1/5).`


`x_1 approx 11.5` °, `x_2 approx 168.5` °.

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