Chapter 5.3, Problem 67E

To determine

Tofind:Thegeneral solution of the equation ${\sec }^{2}x-4\sec x=0$ .

Answer to Problem 67E

Thegeneral solution of the equation ${\sec }^{2}x-4\sec x=0$ is $x={\cos }^{-1}\left(\frac{1}{4}\right)+2n\pi$ .

Explanation of Solution

Given information:

The given equation is ${\sec }^{2}x-4\sec x=0$ .

Calculation:

Solve the given equation.

  $\begin{array}{c}{\sec }^{2}x-4\sec x=0\\ \sec x\left(\sec x-4\right)=0\end{array}$

Equate the factors to zero and solve for $x$ .

  $\sec x=0$

Since, the range of secant is $x\le -1$ and $x\ge 1$ . 0 does not lie in this range. So, there is no

Solve the other factor.

  $\begin{array}{c}\sec x-4=0\\ \sec x=4\\ \cos x=\frac{1}{4}\\ x={\cos }^{-1}\left(\frac{1}{4}\right)+2n\pi \end{array}$

The general solution is

  $x={\cos }^{-1}\left(\frac{1}{4}\right)+2n\pi$

Therefore, thegeneral solution of the equation ${\sec }^{2}x-4\sec x=0$ is $x={\cos }^{-1}\left(\frac{1}{4}\right)+2n\pi$ .