Chapter 7.4, Problem 35E

To determine

To find:The solution of the equation ${\tan }^2\theta -4=0$ .

Answer to Problem 35E

The solutions of the equation ${\tan }^2\theta -4=0$ are $\theta =-1.1+k\pi$ and $\theta =1.1+k\pi$ .

Explanation of Solution

Given information:

The given equation is ${\tan }^2\theta -4=0$ .

Calculation:

Simplify the equation

  $\begin{array}{l}{\tan }^2\theta -4=0\\ {\tan }^2\theta =4\\ \tan \theta =\pm 2\end{array}$

As tangent has period $\pi$ . So first find the solutions in $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ .

Therefore, the solution to the equation in $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ are

  $\begin{array}{c}\theta ={\tan }^{-1}\left(2\right)\\ \approx 1.1\end{array}$

And

  $\begin{array}{c}\theta ={\tan }^{-1}\left(-2\right)\\ \approx -1.1\end{array}$

As tangent repeats its values after every $\pi$ units.

So, the general solution of the equation is given by

  $\theta =1.1+k\pi ,-1.1+k\pi$

Therefore, the solutions of the equation ${\tan }^2\theta -4=0$ are $\theta =-1.1+k\pi$ and $\theta =1.1+k\pi$ .