Chapter 7.5, Problem 29E

(a)

To determine

To find: The solution of the equation $\sec \theta -\tan \theta =\cos \theta$ .

Answer to Problem 29E

The solutions of the equation $\sec \theta -\tan \theta =\cos \theta$ are $\theta =k\pi$ .

Explanation of Solution

Given information:

The given equation is $\sec \theta -\tan \theta =\cos \theta$ .

Calculation:

Simplify the equation.

  $\begin{array}{c}\sec \theta -\tan \theta =\cos \theta \\ \frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }=\cos \theta \\ \frac{1-\sin \theta }{\cos \theta }=\cos \theta \\ 1-\sin \theta ={\cos }^2\theta \end{array}$

Further Simplify.

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

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

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

Now substitute the factor $\sin \theta -1=0$ .

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

  $\begin{array}{c}\theta ={\sin }^{-1}\left(1\right)\\ =\frac{\pi }{2}\end{array}$

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

So, the general solution of the equation is given by

  $\theta =\frac{\pi }{2}+2k\pi$

As the values $\theta =\frac{\pi }{2}+2k\pi$ doesn’t satisfy the equation.

Therefore, the solutions of the equation $\sec \theta -\tan \theta =\cos \theta$ are $\theta =k\pi$ .

(b)

To determine

To find: The solution of the equation $\sec \theta -\tan \theta =\cos \theta$ in $\left[0,2\pi \right)$ .

Answer to Problem 29E

The solution of the equation $\sec \theta -\tan \theta =\cos \theta$ in $\left[0,2\pi \right)$ is $\theta =0,\pi$ .

Explanation of Solution

Given information:

The given equation is $\sec \theta -\tan \theta =\cos \theta$ and the interval is $\left[0,2\pi \right)$ .

Calculation:

According to solution in part(a) the general solution of the equation $\sec \theta -\tan \theta =\cos \theta$ are $\theta =k\pi$ . The solution of the equation in $\left[0,2\pi \right)$ corresponds to $k=0,1$ i.e. $\theta =0,\pi$ .

Therefore, the solution of the equation $\sec \theta -\tan \theta =\cos \theta$ in $\left[0,2\pi \right)$ is $\theta =0,\pi$ .