Chapter 7.5, Problem 5E

To determine

To solve : the given equation $\{{\text{ tan}}^2\theta -2\sec \theta =2$ .

Answer to Problem 5E

The general solutions of $\theta$ for the equation $\{{\text{ tan}}^2\theta -2\sec \theta =2$ are

  $\theta =1.231+2n\pi \text{ , }\theta =5.052+2n\pi \text{ or }\theta =\pi +2n\pi$ .

Explanation of Solution

Given information:

The equation to be solved is $\{{\text{ tan}}^2\theta -2\sec \theta =2$ .

Calculation :

  ${\text{tan}}^2\theta -2\sec \theta =2$............(1)

The trigonometric identity is $\{{\tan }^2\theta ={\sec }^2\theta -1$ .

Put the value of ${\tan }^2\theta$ in equation 1.

  $\begin{array}{l}{\sec }^2\theta -1-2\sec \theta =2\\ {\sec }^2\theta -2\sec \theta -3=0\end{array}$

Put $\sec \theta =x$ .

  $x^2-2x-3=0$

Factorize the above equation.

  $\begin{array}{l}{x}^2-3x+x-3=0\\ x\left(x-3\right)+1\left(x-3\right)=0\\ \left(x+1\right)\left(x-3\right)=0\\ x=-1\text{ or }x=3\end{array}$

Or

  $\begin{array}{l}\sec \theta =-1\text{ or sec}\theta =3\\ or\\ \cos \theta =-1\text{ or cos}\theta =\frac{1}{3}\end{array}$

Now, there are two cases. Solve each case to get the solution of $\theta$ .

Case 1:

  $\cos \theta =-1$

Above equation is satisfied in the interval $\left[0,2\pi \right]$ when $\theta =\pi$ .

Since, the period of cosine function is $2\pi$ . Add $2\pi n$ to get the general solution of $\theta$ where $n$ is any integer.

  $\theta =\pi +2n\pi$

Case 2:

  $\cos \theta =\frac{1}{3}$

Above equation is satisfied in the interval $\left[0,2\pi \right]$ .

Use calculator to solve $\cos \theta =\frac{1}{3}$ .

  $\theta =1.231\text{ radian or }\theta =5.052\text{ radian}$

Since, the period of cosine function is $2\pi$ . Add $2\pi n$ to get the general solution of $\theta$ where $n$ is any integer.

  $\theta =1.231+2n\pi \text{ or }\theta =5.052+2n\pi$

Therefore, the general solutions of $\theta$ for the equation $\{{\text{ tan}}^2\theta -2\sec \theta =2$ are

  $\theta =1.231+2n\pi \text{ , }\theta =5.052+2n\pi \text{ or }\theta =\pi +2n\pi$ .