Chapter 7.4, Problem 54E

To determine

To find: The solution of the equation $\tan \theta \sin \theta +\sin \theta =0$ .

Answer to Problem 54E

The solution of the equation $\tan \theta \sin \theta +\sin \theta =0$ is $\theta =\frac{\pi }{2}+k\pi$ .

Explanation of Solution

Given information:

The given equation is $\tan \theta \sin \theta +\sin \theta =0$ .

Calculation:

Simplify the equation.

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

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

As sine 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

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

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

So, the general solution of the equation is given by

  $\begin{array}{c}\theta =0+2k\pi ,\pi +2k\pi \\ =0+k\pi \\ =k\pi \end{array}$

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

Now substitute the factor $\tan \theta +1=0$ .

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

  $\begin{array}{c}\theta ={\tan }^{-1}\left(-1\right)\\ =-\frac{\pi }{4}\end{array}$

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

So, the general solution of the equation is given by

  $\theta =-\frac{\pi }{4}+k\pi$

Therefore, the solutions of the equation $\tan \theta +1=0$ are $\theta =-\frac{\pi }{4}+k\pi$ .

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