Chapter 7.5, Problem 30E

(a)

To determine

To find: The solution of the equation $\tan 3\theta +1=\sec 3\theta$ .

Answer to Problem 30E

The solutions of the equation $\tan 3\theta +1=\sec 3\theta$ are $\theta =\frac{2k\pi }{3}$ .

Explanation of Solution

Given information:

The given equation is $\tan 3\theta +1=\sec 3\theta$ .

Calculation:

Simplify the equation.

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

According to identity.

  ${\sec }^{2}3\theta -{\tan }^{2}3\theta =1$

Further Simplify.

  $\begin{array}{c}1+2\tan 3\theta =1\\ 2\tan 3\theta =0\\ \tan 3\theta =0\end{array}$

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

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

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

So, the general solution of the equation is given by

  $\begin{array}{c}3\theta =0+k\pi \\ \theta =\frac{k\pi }{3}\end{array}$

As the odd multiples of solution doesn’t satisfy the equation.

Therefore, the solutions of the equation $\tan 3\theta +1=\sec 3\theta$ are $\theta =\frac{2k\pi }{3}$ .

(b)

To determine

To find: The solution of the equation $\tan 3\theta +1=\sec 3\theta$ in $\left[0,2\pi \right)$ .

Answer to Problem 30E

The solution of the equation $\tan 3\theta +1=\sec 3\theta$ in $\left[0,2\pi \right)$ is $\theta =0,\frac{2\pi }{3},\frac{4\pi }{3}$ .

Explanation of Solution

Given information:

The given equation is $\tan 3\theta +1=\sec 3\theta$ 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 $\theta =\frac{2k\pi }{3}$ . The solution of the equation in $\left[0,2\pi \right)$ corresponds to $k=0,1,2$ i.e. $\theta =0,\frac{2\pi }{3},\frac{4\pi }{3}$ .

Therefore, the solution of the equation $\tan 3\theta +1=\sec 3\theta$ in $\left[0,2\pi \right)$ is $\theta =0,\frac{2\pi }{3},\frac{4\pi }{3}$ .