Chapter 5.4, Problem 70E

To determine

To calculate: The solution of the given equation in the given interval.

Answer to Problem 70E

The solution of the given equation in the interval $\text{[0,2}\pi \text{)}$ is $\frac{3\pi }{2}$ .

Explanation of Solution

Given information:

The following equation:

  $\cos \left(x+\frac{\pi }{6}\right)-\cos \left(x-\frac{\pi }{6}\right)=1$

The following interval:

  $\text{[0,2}\pi \text{)}$

Formula used:

The following sum formula:

  $\cos (u+v)=\cos u\cos v-\sin u\sin v$

The following difference formula:

  $\cos (u-v)=\cos u\cos v+\sin u\sin v$

Calculation:

The given equation can be solved as:

  $\begin{array}{l}\cos \left(x+\frac{\pi }{6}\right)-\cos \left(x-\frac{\pi }{6}\right)=1\text{ [Given]}\\ \left\{\cos x\cos \frac{\pi }{6}-\sin x\sin \frac{\pi }{6}\right\}-\left\{\cos x\cos \frac{\pi }{6}+\sin x\sin \frac{\pi }{6}\right\}\text{=1 [Sum and difference formula]}\\ \cos x\cos \frac{\pi }{6}-\sin x\sin \frac{\pi }{6}-\cos x\cos \frac{\pi }{6}-\sin x\sin \frac{\pi }{6}\text{=1 [Simplify]}\\ -2\sin x\sin \frac{\pi }{6}=1\text{ [Add]}\\ -2\cdot \frac{1}{2}\sin x=\text{1 [Put value of sin(}\pi /6)]\\ -\sin x=1\text{ [Simplify]}\\ x=\arcsin (-1)\text{ [Inverse trigonometric function]}\\ x=\frac{3\pi }{2}\text{ [Value of arcsin(-1)] }\end{array}$

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 $\cos \left(x+\frac{\pi }{6}\right)-\cos \left(x-\frac{\pi }{6}\right)$ which is required to graph.

Step 4: Press GRAPH button to graph the function.

The graph is obtained as:

  

Interpretation:

In the interval $\text{[0,2}\pi \text{)}$ , the value of the function is $\text{1}$ at $\frac{3\pi }{2}$ . Hence, verified.