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Chapter 6: Problem 38

Find the exact solutions of the given equations, in radians, that lie in the interval \([0,2 \pi)\). $$\sin x-\csc x=0$$

Short Answer

Expert verified
The exact solutions of the equation \(\sin x - \csc x = 0\) that lie in the interval [0, 2pi) are \(x = \pi/2\) and \(x = 3\pi/2\).

Step by step solution

01

Simplification using Trigonometric Identity

The \(\csc\) function is the reciprocal of the \(\sin\) function. Hence, we can rewrite \(\csc x\) as \(1/ \sin x\). So the equation \(\sin x - \csc x = 0\) becomes \(\sin x - 1/ \sin x = 0\).
02

Solving the Equation

To make the equation easier to solve, multiply all the parts of the equation by \(\sin{x}\) to eliminate the fraction. This will result in \(\sin^2 x - 1 = 0\). Re-arranging, we get \(\sin^2 x = 1\). Taking the square root on both sides, we find the solutions to be \(\sin x = 1\) and \(\sin x = -1\).
03

Determining the solutions in the interval [0, 2pi)

We now need to find the solutions of \(\sin x = 1\) and \(\sin x = -1\) within the interval \([0, 2\pi)\). From the unit circle or using a calculator, we can see that \(\sin x = 1\) at \(x = \pi/2\). The equation \(\sin x = -1\) gives \(x = \frac{3 \pi}{2}\). Hence, the solutions of the equation \(\sin x - \csc x = 0\) that lie in the interval \([0,2 \pi)\) are \(x = \pi/2\) and \(x = 3 \pi/2\).

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