Problem 60 - Precalculus: Functions and Graphs

Question

Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$\sec ^{5} \theta=4 \sec \theta$$

Step-by-Step Solution

Answer

\(\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\).

Step 1: Simplify the Equation

The given equation is \(\sec^{5} \theta = 4 \sec \theta\). First, rearrange it into a polynomial form by subtracting 4\(\sec \theta\) from both sides, resulting in \(\sec^{5} \theta - 4 \sec \theta = 0\).

Step 2: Factor Out Common Terms

Next, factor out \(\sec \theta\) from the expression: \(\sec \theta (\sec^{4} \theta - 4) = 0\). This gives us two potential solutions: \(\sec \theta = 0\) or \(\sec^{4} \theta - 4 = 0\).

Step 3: Solve \(\sec \theta = 0\)

The secant function, \(\sec \theta\), is equal to \(\frac{1}{\cos \theta}\). So, \(\sec \theta = 0\) implies \(\cos \theta = \frac{1}{0}\), which is undefined. Therefore, there are no solutions for \(\sec \theta = 0\).

Step 4: Solve \(\sec^{4} \theta - 4 = 0\)

Set \(\sec^{4} \theta = 4\), giving \(\sec^2 \theta = 2\) after taking the square root. Thus, \(\cos^2 \theta = \frac{1}{2}\) (since \(\sec \theta = \frac{1}{\cos \theta}\)).

Step 5: Solve \(\cos^2 \theta = \frac{1}{2}\)

Take the square root of both sides: \(\cos \theta = \pm \frac{1}{\sqrt{2}}\). Which simplifies to \(\cos \theta = \pm \frac{\sqrt{2}}{2}\).

Step 6: Find Angles for \(\cos \theta = \frac{\sqrt{2}}{2}\)

\(\cos \theta = \frac{\sqrt{2}}{2}\) corresponds to angles \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{7\pi}{4}\).

Step 7: Find Angles for \(\cos \theta = -\frac{\sqrt{2}}{2}\)

\(\cos \theta = -\frac{\sqrt{2}}{2}\) corresponds to angles \(\theta = \frac{3\pi}{4}\) and \(\theta = \frac{5\pi}{4}\).

Key Concepts

Secant Function

The secant function, often written as \(\sec \theta\), is a key concept in trigonometry. It is defined as the reciprocal of the cosine function, hence \(\sec \theta = \frac{1}{\cos \theta}\). This means that wherever the cosine value is zero, the secant function becomes undefined, as division by zero is not permissible. The secant function exhibits its own set of unique values and behaviors.
Understanding how secant relates to cosine is crucial when solving trigonometric equations. For instance, if you're given an equation like \(\sec^{5} \theta = 4 \sec \theta\), it is helpful to translate it in terms of cosine. Converting between these related functions allows for easier manipulation and understanding of the problem.
You'll often begin by recognizing places where the equation is undefined (in this case, where \(\cos \theta = 0\)), and then proceed to find meaningful solutions within a given interval.

Interval Notation

Interval notation is a concise way to describe a segment of the number line. It is especially useful in specifying where solutions are valid in trigonometric contexts—like indicating the permissible range for \(\theta\) in trigonometric equations.
In this problem, we look for solutions in the interval \([0, 2\pi)\). This notation means that we include 0, but not \(2\pi\). Such intervals are crucial when looking at cyclic functions like sine, cosine, and secant, as they repeat every \(2\pi\) radians. This interval covers one full period of these functions, ensuring that we account for every possible value without repetition.
Utilizing interval notation allows mathematicians to clearly express which range of values solutions should be derived from. It's a key form of communication in mathematics, providing precision and clarity.

Factoring Equations

Factoring is a powerful method to simplify trigonometric equations. It helps to break down complex items into more manageable units. Let's consider \(\sec^{5}\theta = 4\sec\theta\). By moving all terms to one side, we form the expression \(\sec^{5} \theta - 4 \sec \theta = 0\). Factoring allows us to rewrite this as \(\sec \theta (\sec^{4} \theta - 4) = 0\).
Factoring is key to identifying a trigonometric equation's solutions. After factoring, we have each individual factor set to zero. From this, we discern separate equations—such as \(\sec \theta = 0\) or \(\sec^{4} \theta - 4 = 0\). These can be approached individually to find specific solutions.
Understanding factoring prepares you to solve increasingly complicated problems, as it often lets you reduce a polynomial to its simplest form and therefore get more accessible solutions.

Cosine Values

In trigonometry, understanding how to derive and apply cosine values is essential for solving equations. The cosine function, defined for any angle \(\theta\), is a ratio that comes from the unit circle. For any angle, \(\cos \theta = \frac{ ext{adjacent side}}{ ext{hypotenuse}}\) in terms of a right triangle relationship.
In this trigonometric problem, we need to find angles where the cosine function equals certain values, specifically \(\cos \theta = \pm \frac{\sqrt{2}}{2}\). These values correspond to standard angles on the unit circle. For \(\cos \theta = \frac{\sqrt{2}}{2}\), the corresponding angles in our interval are \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{7\pi}{4}\). Conversely, for \(\cos \theta = -\frac{\sqrt{2}}{2}\), the angles are \(\theta = \frac{3\pi}{4}\) and \(\theta = \frac{5\pi}{4}\).
Knowing exact cosine values for key angles is a fundamental skill. It allows you to quickly find solutions to trigonometric equations and validates your solution within the specified interval.