Chapter 5.3, Problem 32E

To determine

To solve: ${\tan }^{2}x=\sec x-1$ in interval $\left[0,2\pi \right)$ .

Answer to Problem 32E

The solution of equation is $\pi \in \left[0,2\pi \right)$ .

Explanation of Solution

Given:

The equation is ${\tan }^{2}x=\sec x-1$ .

Calculation:

Solve the trigonometry equation as follows:

  $\begin{array}{l}{\tan }^{2}x=\sec x-1\left({\sec }^{2}x=1+{\tan }^{2}x\right)\\ {\sec }^{2}x-1=\sec x-1\\ {\sec }^{2}x-\sec x=0\\ \sec x\left(\sec x-1\right)=0\\ \sec x=0\text{and}\sec x=1\\ \text{No}\text{solution}\text{exist}\text{}\text{for}\sec x=0\\ \text{For}\sec x=1,\\ x=\pi \in \left[0,2\pi \right)\end{array}$

Thus, the solution of equation is $\pi \in \left[0,2\pi \right)$ .