Chapter 7.4, Problem 50E

To determine

To find: The solution of the equation $3{\tan }^3\theta =\tan \theta$ .

Answer to Problem 50E

The solutions of the equation $3{\tan }^3\theta =\tan \theta$ are $\theta =k\pi$ , $\theta =-\frac{\pi }{6}+k\pi$ and $\theta =\frac{\pi }{6}+k\pi$ .

Explanation of Solution

Given information:

The given equation is $3{\tan }^3\theta =\tan \theta$ .

Calculation:

Simplify the given equation.

  $\begin{array}{c}3{\tan }^3\theta =\tan \theta \\ 3{\tan }^3\theta -\tan \theta =0\\ \tan \theta \left(3{\tan }^2\theta -1\right)=0\end{array}$

First substitute the factor $\tan \theta =0$ .

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

  $\begin{array}{c}\theta ={\tan }^{-1}\left(0\right)\\ =0\end{array}$

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

So, the general solution of the equation is given by

  $\begin{array}{c}\theta =0+k\pi \\ =k\pi \end{array}$

Therefore, the solutions of the equation $\tan \theta =0$ are $\theta =k\pi$ .

Now substitute the factor $3{\tan }^2\theta -1=0$ .

Simplify the equation.

  $\begin{array}{c}3{\tan }^2\theta -1=0\\ 3{\tan }^2\theta =1\\ {\tan }^2\theta =\frac{1}{3}\\ \tan \theta =\pm \frac{1}{\sqrt{3}}\end{array}$

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

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ =\frac{\pi }{6}\end{array}$

And,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)\\ =-\frac{\pi }{6}\end{array}$

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

So, the general solution of the equation is given by

  $\theta =-\frac{\pi }{6}+k\pi ,\frac{\pi }{6}+k\pi$

Therefore, the solutions of the equation $3{\tan }^2\theta -1=0$ are $\theta =-\frac{\pi }{6}+k\pi$ and $\theta =\frac{\pi }{6}+k\pi$ .

Therefore, the solutions of the equation $3{\tan }^3\theta =\tan \theta$ are $\theta =k\pi$ , $\theta =-\frac{\pi }{6}+k\pi$ and $\theta =\frac{\pi }{6}+k\pi$ .