Problem 72 - Precalculus with Limits

Question

In Exercises 63 - 74, use inverse functions where needed to find all solutions of the equation in the interval \( [0,2\pi) \). \( \sec^2 x + 2 \sec x - 8 = 0 \)

Step-by-Step Solution

Answer

Therefore, the solutions to the equation in the interval \( [0,2\pi) \) are \( x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} \).

Step 1: Convert secant to cosine

Since secant of x equals the reciprocal of cosine of x, replace secant with its counterpart, you will have the expression as : \( (\frac{1}{\cos x})^2 + 2(\frac{1}{\cos x}) - 8 = 0 \).

Step 2: Simplify to standard form of quadratic equation

Multiply the entire equation by \( \cos^2x \) to remove the denominator: \(1+ 2 \cos x - 8\cos^2 x = 0 \)

Step 3: Rewrite as quadratic

Let's rewrite this in the standard form of a quadratic equation \(ax^2 + bx + c = 0\). Here \(a = -8, b = 2, and c = 1\). So the equation is: \( -8\cos^2x + 2\cos x + 1 = 0\).

Step 4: Solve the quadratic equation

The quadratic formula is: \( x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\). Substituting the same, \( \cos x = \frac{-2\pm \sqrt{4-4(-8)(1)}}{2(-8)} = \frac{-2\pm \sqrt{36}}{-16} = \frac{-2\pm 6}{-16}.\) So, we have two solutions, \( \cos x = \frac{1}{2} \) and \( \cos x = \frac{-1}{2} \).

Step 5: Find \(x\) from cos values

From the solution in Step 4, we find the values of x within the given interval. Considering \( \cos x = \frac{1}{2} \), we have x equals \( \frac{\pi}{3} \) and \( \frac{5\pi}{3} \), since cosine function equals \( \frac{1}{2} \) at these angles in the interval \( [0, 2\pi) \). Considering \( \cos x = \frac{-1}{2} \), we have x equals \( \frac{2\pi}{3} \) and \( \frac{4\pi}{3} \), since cosine function equals \( \frac{-1}{2} \) at these angles in the interval \( [0, 2\pi) \).

Key Concepts

Secant

In trigonometry, the secant function, denoted as \( \sec x \), is the reciprocal of the cosine function. This means that \( \sec x = \frac{1}{\cos x} \). Secant and cosine are related, but they have different ranges and domains.

Important to know about secant:

• Secant is undefined wherever cosine is zero because division by zero is not possible.
• The secant function is periodic with period \(2\pi\), similar to cosine.
• It can take values from 1 to infinity or -1 to negative infinity.

In the exercise, secant is converted to cosine as a step to make solving the equation more straightforward. Using the reciprocal relationship between secant and cosine allows us to rewrite secant functions in terms of cosine, which is often easier to work with, especially in algebraic equations.

Cosine

The cosine function is one of the fundamental functions in trigonometry, denoted as \( \cos x \). It is often used in situations involving right triangles and is key to understanding many other trigonometric functions.

What you should know about cosine:

• Cosine measures the adjacent side over the hypotenuse in a right triangle.
• It has values ranging from -1 to 1.
• The function is periodic with period \(2\pi\), meaning it repeats every \(2\pi\) radians.

In our given problem, the secant is expressed in terms of cosine to change the trigonometric equation into a traditional algebraic equation. This transformation allows the use of different algebraic methods, such as factoring or applying the quadratic formula, simplifying the problem-solving process.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation of the form \(ax^2 + bx + c = 0\). To solve these equations, we can use several methods like factoring, completing the square, or the quadratic formula.

Key points about quadratic equations:

• They often have two solutions, which can be real or complex numbers.
• The discriminant, given by \(b^2 - 4ac\), determines the nature of the roots.
• The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), used when factoring isn't feasible.

In our exercise, the original trigonometric equation was transformed into a quadratic form by substituting \(\sec x\) with \(\cos x\) and clearing the fractions. Once in standard quadratic form, we applied the quadratic formula to find the cosine values, which allowed us to identify possible angles \(x\) within the specified interval. This showcases how a trigonometric problem can be tackled using concepts from algebra.