Chapter 7.4, Problem 44E

To determine

To find:The solution of the equation ${\tan }^4\theta -13{\tan }^2\theta +36=0$ .

Answer to Problem 44E

The solutions of the equation ${\tan }^4\theta -13{\tan }^2\theta +36=0$ are $\theta =-1.11+k\pi$ , $\theta =1.11+k\pi$ , $\theta =-1.25+k\pi$ and $\theta =1.25+k\pi$ .

Explanation of Solution

Given information:

The given equation is ${\tan }^4\theta -13{\tan }^2\theta +36=0$ .

Calculation:

Simplify the given equation.

  $\begin{array}{c}{\tan }^4\theta -13{\tan }^2\theta +36=0\\ {\tan }^4\theta -9{\tan }^2\theta -4{\tan }^2\theta +36=0\\ \left({\tan }^2\theta -4\right)\left({\tan }^2\theta -9\right)=0\end{array}$

First substitute the factor ${\tan }^2\theta -4=0$ .

Simplify the equation

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

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

  $\begin{array}{c}\theta ={\tan }^{-1}\left(2\right)\\ \approx 1.11\end{array}$

And,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(-2\right)\\ \approx -1.11\end{array}$

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

So, the general solution of the equation is given by

  $\theta =-1.11+k\pi ,1.11+k\pi$

Therefore, the solutions of the equation ${\tan }^2\theta -4=0$ are $\theta =-1.11+k\pi$ and $\theta =1.11+k\pi$ .

Now substitute the factor ${\tan }^2\theta -9=0$ .

Simplify the equation

  $\begin{array}{c}{\tan }^2\theta -9=0\\ {\tan }^2\theta =9\\ \tan \theta =\pm 3\end{array}$

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

  $\begin{array}{c}\theta ={\tan }^{-1}\left(3\right)\\ \approx 1.25\end{array}$

And,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(-3\right)\\ \approx -1.25\end{array}$

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

So, the general solution of the equation is given by

  $\theta =-1.25+k\pi ,1.25+k\pi$

Therefore, the solutions of the equation ${\tan }^2\theta -9=0$ are $\theta =-1.25+k\pi$ and $\theta =1.25+k\pi$ .

Combining all the solutions give the solution of the equation ${\tan }^4\theta -13{\tan }^2\theta +36=0$ .

Therefore, the solutions of the equation ${\tan }^4\theta -13{\tan }^2\theta +36=0$ are $\theta =-1.11+k\pi$ , $\theta =1.11+k\pi$ , $\theta =-1.25+k\pi$ and $\theta =1.25+k\pi$ .