Chapter 7.4, Problem 37E

To determine

To find: The solution of the equation ${\sec }^2\theta -2=0$ .

Answer to Problem 37E

The solutions of the equation ${\sec }^2\theta -2=0$ are $\theta =\frac{\pi }{4}+k\pi$ and $\theta =\frac{3\pi }{4}+k\pi$ .

Explanation of Solution

Given information:

The given equation is ${\sec }^2\theta -2=0$ .

Calculation:

Simplify the equation

  $\begin{array}{c}{\sec }^2\theta -2=0\\ {\sec }^2\theta =2\\ \sec \theta =\pm \sqrt{2}\end{array}$

As secant has period $2\pi$ . So first find the solutions in $\left[0,2\pi \right)$ .

Therefore, the solution to the equation in $\left[0,2\pi \right)$ are

  $\begin{array}{c}\theta ={\sec }^{-1}\left(\sqrt{2}\right)\\ =\frac{\pi }{4},\frac{7\pi }{4}\end{array}$

And

  $\begin{array}{c}\theta ={\sec }^{-1}\left(-\sqrt{2}\right)\\ =\frac{3\pi }{4},\frac{5\pi }{4}\end{array}$

As secant repeats its values after every $2\pi$ units.

So, the general solution of the equation is given by

  $\theta =\frac{\pi }{4}+2k\pi ,\frac{3\pi }{4}+2k\pi ,\frac{5\pi }{4}+2k\pi ,\frac{7\pi }{4}+2k\pi$

Therefore, the solutions of the equation ${\sec }^2\theta -2=0$ are $\theta =\frac{\pi }{4}+k\pi$ and $\theta =\frac{3\pi }{4}+k\pi$ .