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Chapter 5: Problem 73

Use inverse functions where needed to find all solutions of the equation in the interval \([0,2 \pi)\). $$\csc ^{2} x+3 \csc x-4=0$$

Short Answer

Expert verified
The solution to the exercise equation \( \csc ^{2} x+3 \csc x-4=0 \) in the interval [0,2蟺) is \( x = 蟺/2 \)

Step by step solution

01

Identify the Form of the Equation

The equation \( \csc ^{2} x+3 \csc x-4=0 \) is a quadratic in terms of \( \csc x \). We can solve it similarly to other quadratics. This quadratic form is recognizable as \( ax^2 + bx - c = 0 \), with \( a = 1, b = 3, c = -4 \), and where \(x = \csc x\).
02

Factor the Quadratic

By factoring the quadratic equation, we get \( (\csc x + 4)(\csc x - 1) = 0 \)
03

Nullify Each Factor

Next, set each factor equal to zero to find the values of \( x \). This gives us \( \csc x + 4 = 0 \) and \( \csc x - 1 = 0 \), yielding \( \csc x = -4 \) and \( \csc x = 1 \), respectively.
04

Solve for x

Now, remember that \( \csc x = 1/sin x \). So we are looking for angles \(x\) such that \(sin x = -1/4\) or \(sin x = 1\). Referring to the unit circle or a sine graph, we find that:For \( sin x = -1/4 \), there are no solutions in the interval [0,2蟺).For \( sin x = 1 \), \( x = 蟺/2 \) in the interval [0,2蟺)
05

Conclusion

After all these steps, the conclusion can be drawn that the unique solution for \( x \) in the inverval [0, 2蟺) is \( x = 蟺/2 \)

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