Chapter 7.5, Problem 62E

To determine

To find: The solution of the equation $\cos x=\frac{1}{2}\left(e^x+e^{-x}\right)$ , rounded to two decimal places.

Answer to Problem 62E

The solution of the equation $\cos x=\frac{1}{2}\left(e^x+e^{-x}\right)$ is $x=0$ .

Explanation of Solution

Given information:

The given equation is $\cos x=\frac{1}{2}\left(e^x+e^{-x}\right)$ .

Calculation:

Using graphing device.

Let’s plot graph for the two functions. The left hand side is displayed in blue, while the right hand side is in purple.

The graph is shown in figure (1).

  

Figure (1)

The intersection (the solution) are indicated by red dots. So, there are one visible solutions.

  $x=0$

To graph of two functions on a TI calculator, first press the $\text{Y}=$ button. You will see a list of places you can input functions. To the right of $"\backslash Y_1"$ , type $"\sin (2x)"$ , and type $"x"$ next to $"\backslash Y_2"$ . Then press $\text{"GRAPH"}$ button. The two functions should be displayed on a $\text{10}$ by $\text{10}$ coordinate plane.

In order to find the intersections of the two functions (which represent the solutions to the equation that equate them), first press the $"2\text{nd"}$ button, and then the $\text{"TRACE/CALC"}$ button. Arrow down to the option that says $\text{"intersects"}$ and press $\text{"ENTER"}$ . It should bring you to the graph screen.

It will say $\text{"First}\text{curve?"}$ at this point, use the left and right arrow to move the cursor just to the left of the intersection you want to identify and press enter. It will say $\text{"Second}\text{curve?"}$ . Now, use the left and right arrow to move the cursor just to the right of the intersection you want to identify and press enter. It will say $\text{"Guess"}$ press enter. It will say $\text{"intersection"}$ , along with $\text{"x}="$ and $\text{"y}="$ values.

The $x$ value is the solution. Do this for every intersection you see on the graph, and will have found every solution to the equation.

Therefore, the solution of the equation $\cos x=\frac{1}{2}\left(e^x+e^{-x}\right)$ is $x=0$ .