Chapter 5.5, Problem 68E

To determine

To find: The solutions of the given equation in the given interval.

Answer to Problem 68E

The solutions of the given equation in the interval $[0,2\pi )$ are $x=0,\frac{\pi }{3},\frac{5\pi }{3}$ .

Explanation of Solution

Given information:

The following equation:

  $\sin \frac{x}{2}+\cos x-1=0$

The following interval:

  $[0,2\pi )$

Formula used:

The following double angle formula

  $\cos 2u=1-2{\sin }^{2}u$

Calculation:

The given equation can be solved as follows:

  $\begin{array}{l}\sin \frac{x}{2}+\cos x-1=0\text{ [Given]}\\ \sin \frac{x}{2}+\cos 2\left(\frac{x}{2}\right)-1=0\text{ [Multiply and divide by 2]}\\ \sin \frac{x}{2}+1-2{\sin }^2\frac{x}{2}-1=0\text{ [Double angle formula]}\\ \sin \frac{x}{2}-2{\sin }^2\frac{x}{2}=0\text{ [Add like terms]}\\ \sin \frac{x}{2}\left(1-2\sin \frac{x}{2}\right)=0\text{ [Factor]}\\ \sin \frac{x}{2}=0\text{ 1}-2\sin \frac{x}{2}=0\text{ [Set each factor to 0]}\\ \sin \frac{x}{2}=0\text{ sin}\frac{x}{2}=\frac{1}{2}\text{ [Simplify]}\\ \frac{x}{2}=\arcsin \left(0\right)\frac{x}{2}=\arcsin \left(\frac{1}{2}\right)\text{ [Inverse trigonometric function]}\\ \frac{x}{2}=0,\pi ,2\pi \frac{x}{2}=\frac{\pi }{6}\text{,}\frac{5\pi }{6}\text{ [Find values of inverse]}\\ x=0,2\pi ,4\pi x=\frac{\pi }{3}\text{,}\frac{5\pi }{3}\text{ [Multiply both sides by 2]}\end{array}$

Among all the solutions of $x$ , $x=0,\frac{\pi }{3},\frac{5\pi }{3}$ lie in the given interval. The solution is verified graphically as

Graph:

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $\sin \frac{x}{2}+\cos x-1$ and which is required to graph.

Step 4: Press GRAPH button to graph the function.

The graph is obtained as:

  

Interpretation:

In the interval $[0,2\pi )$ , the value of the equation is $0$ at $x=0,\frac{\pi }{3},\frac{5\pi }{3}$ . Hence verified.