Chapter 5.3, Problem 80E

To determine

To solve: The given equation using inverse trigonometric functions.

Answer to Problem 80E

The solutions of the equation are:

  $\begin{array}{l}x=\arctan (-2)+n\pi \\ x=\frac{\pi }{4}+n\pi \end{array}$

Explanation of Solution

Given information:

The following equation:

  ${\sec }^{2}x+\tan x-3=0$

Formula used:

The following identity:

  $1+{\tan }^{2}x={\sec }^{2}x$

Calculation:

The equation is solved as follows:

  $\begin{array}{l}{\sec }^{2}x+\tan x-3=0\\ 1+{\tan }^{2}x+\tan x-3=0\text{ [Pythagorean identity]}\\ {\tan }^{2}x+\tan x-2=0\text{ [Combine like terms]}\\ (\tan x+2)(\tan x-1)=0\text{ [Factor]}\end{array}$

Setting each factor equal to zero, two solutions are obtained in the interval $(-\pi /2,\pi /2)$

[Range of inverse tangent function is $(-\pi /2,\pi /2)$ ]

  $\begin{array}{l}\tan x+2=0\\ \tan x=-2\text{ [Subtracting 2 from both sides]}\\ x=\text{arctan(}-2)\text{ [Inverse tangent function]}\end{array}$

and

  $\begin{array}{l}\tan x-1=0\\ \tan x=1\text{ [Adding 1 to both sides]}\\ x=\text{arctan}1\text{ [Inverse tangent function]}\\ x=\frac{\pi }{4}\text{ [Value of arctan}1]\end{array}$

Finally, because $\tan x$ has a period of $\pi$ , add multiples of $\pi$ to obtain,

  $x=\arctan (-2)+n\pi$ and $x=\frac{\pi }{4}+n\pi$ [General solution]

Where $n$ is an integer.

The solution is confirmed graphically as follows:

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 ${\sec }^{2}x+\tan x-3$ which is required to graph.

Step 4: Press GRAPH button to graph the function.

The graph is obtained as: