Chapter 5.3, Problem 84E

(a)

To determine

To find : the maximum and minimum point of the function

Answer to Problem 84E

The maximum points of the function are $\underset{\_}{\boxed{\left(\pi ,-\pi -1\right)}}$ radians.

The minimum point of the function are $\underset{\_}{\boxed{\left(0,1\right)}}$ radians.

Explanation of Solution

Given information : Function $f\left(x\right)=\sec x+\tan x-x$ ; Trigonometric Equation $\sec x\tan x+{\sec }^{2}x=1$ ; $\left[0,2\pi \right)$

Concept Involved:

A maximum is a high point and a minimum is a low point over the given interval.

Graph:

  

Interpretation:

The graph of the function $f\left(x\right)=secx+tanx-x$ shows, at $x=\pi$ radians the graph has the maximum value of $-\pi -1$ and at $x=0$ radians the graph has the minimum value of $1$ .

(b)

To determine

To find : all the solutions of the trigonometric equations in the given interval

Answer to Problem 84E

The solution to the given trigonometric equation are $\underset{\_}{\boxed{x=0,\&\pi }}$ radians.

Explanation of Solution

Given information : Function $f\left(x\right)=\sec x+\tan x-x$ ; Trigonometric Equation $\sec x\tan x+{\sec }^{2}x=1$ ; $\left[0,2\pi \right)$

Concept Involved:

Solution to a quadratic equation are the values of x that makes the equation TRUE.

To solve a trigonometric equation, use standard algebraic techniques (when possible) such as collecting like terms, extracting square roots, and factoring.

Our preliminary goal in solving a trigonometric equation is to isolate the trigonometric function on one side of the equation.

Calculation:

Subtracting 1 on both sides of the equation $\sec x\tan x+{\sec }^{2}x=1$

  $\sec x\tan x+{\sec }^{2}x-1=1-1$

Simplify on both sides of the equation

  $\sec x\tan x+{\sec }^{2}x-1=0$

Use the Pythagorean Identity ${\sec }^{2}x-1={\tan }^{2}x$

  $\sec x\tan x+{\tan }^{2}x=0$

Factor the Greatest Common Factor in the left side of the equation

  $\tan x\left(\sec x+\tan x\right)=0$

Using the zero factor property which states that if $a\cdot b=0$ then $a=0\left(or\right)b=0$ , we need to set each factor to zero.

  $\tan x=0$▶1^st equation

  $\sec x+\tan x=0$▶2^nd equation

Solving the 1^st equation and finding x values that makes it true in the interval $\left[0,2\pi \right)$

• By taking inverse tangent on both sides
${\tan }^{-1}\left(\tan x\right)=0$

  $x=0,\pi$

Solving the 2^nd equation and finding x values that makes it true in the interval $\left[0,2\pi \right)$

• By rewriting left side of the equation using reciprocal identity and quotient identity

  $\frac{1}{\cos x}+\frac{\sin x}{\cos x}=0$

  $\frac{1+\sin x}{\cos x}=0$

At $x=\frac{\pi }{2}$ , the numerator is zero, but it makes the denominator also zero. So there is no solution to this equation.

Conclusion:

  $x=0\&\pi$ radians are the zero of the function and solution to the equation $\sec x\tan x+{\sec }^{2}x=1$ in the interval $\left[0,2\pi \right)$