Problem 37 - Precalculus Mathematics for Calculus

Question

Find all solutions of the given equation. $$\sec ^{2} \theta-2=0$$

Step-by-Step Solution

Answer

\(\theta = \frac{\pi}{4} + 2k\pi, \frac{7\pi}{4} + 2k\pi, \frac{3\pi}{4} + 2k\pi, \frac{5\pi}{4} + 2k\pi\) for integer \(k\).

Step 1: Set up the Equation

We start with the given equation \(\sec^2 \theta - 2 = 0\). Our goal is to solve for the angle \(\theta\) in this equation.

Step 2: Isolate \(\sec^2 \theta\)

To isolate \(\sec^2 \theta\), add 2 to both sides of the equation: \(\sec^2 \theta = 2\).

Step 3: Use Trigonometric Identity

Recall that \(\sec \theta = \frac{1}{\cos \theta}\). We can rewrite \(\sec^2 \theta = 2\) as \(\left(\frac{1}{\cos \theta}\right)^2 = 2\).

Step 4: Solve for \(\cos \theta\)

Taking the reciprocal square root of both sides, we have \(\cos^2 \theta = \frac{1}{2}\). Taking the square root gives us \(\cos \theta = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}\).

Step 5: Find the General Solutions

The solutions to \(\cos \theta = \frac{\sqrt{2}}{2}\) are \(\theta = \frac{\pi}{4} + 2k\pi\) and \(\theta = \frac{7\pi}{4} + 2k\pi\). The solutions to \(\cos \theta = -\frac{\sqrt{2}}{2}\) are \(\theta = \frac{3\pi}{4} + 2k\pi\) and \(\theta = \frac{5\pi}{4} + 2k\pi\), where \(k\) is an integer.

Key Concepts

Secant Function

The secant function is a trigonometric function, which is often abbreviated as "sec". It is an important part of trigonometry, as it relates to the cosine function. The secant of an angle, denoted as \( \sec \theta \), is defined as the reciprocal of the cosine of that angle:

• \( \sec \theta = \frac{1}{\cos \theta} \)

This means that wherever the cosine function is zero, the secant function is undefined, since division by zero is not allowed. The range of \( \sec \theta \) is all real numbers except values between -1 and 1. Understanding how secant connects to cosine is crucial, as seen in solving the equation \( \sec^2 \theta = 2 \). We recognize that the problem involves squaring the secant, leading us to consider the square of the cosine function as its inverse.

Cosine Function

The cosine function, represented as \( \cos \theta \), is one of the primary trigonometric functions. It describes the x-coordinate of a point on the unit circle as it relates to a given angle \( \theta \). This function repeats every \( 2\pi \) radians and has a range from -1 to 1. Since \( \sec \theta \) is the reciprocal of \( \cos \theta \), knowing how to manipulate the cosine function can help us solve complex trigonometric equations.
In the given equation \( \sec^2 \theta = 2 \), converting it to a cosine form, we use the identity \( \sec \theta = \frac{1}{\cos \theta} \). Thus, simplifying gives \( \cos^2 \theta = \frac{1}{2} \). Solving this involves determining values of \( \theta \) where \( \cos \theta = \pm \frac{\sqrt{2}}{2} \), which are common special angles on the unit circle.

Trigonometric Identity

Trigonometric identities are formulas that express relationships between trigonometric functions. One such identity is the Pythagorean identity, which is crucial in deriving relationships like \( \sec^2 \theta = 1 + \tan^2 \theta \). These identities are tools for simplifying equations and deriving solutions.
In the problem at hand, the identity \( \sec \theta = \frac{1}{\cos \theta} \) simplifies \( \sec^2 \theta = 2 \) to \( \cos^2 \theta = \frac{1}{2} \). Recognizing and using this identity is what transforms a more complex equation into a manageable form. Such identities are invaluable for students learning how to solve trigonometric equations, as they reveal the connections between different trigonometric functions.

General Solutions of Trigonometric Equations

Finding general solutions of trigonometric equations involves determining all possible solutions within the set of real numbers. For periodic functions like cosine and secant, solutions form repeating patterns.

• The cosine function has a fundamental period of \(2\pi\), meaning every solution \(\theta\) can be expressed as \(\theta\) plus any integer multiple of \(2\pi\).

In our equation \(\cos \theta = \pm \frac{\sqrt{2}}{2} \), we find specific angle solutions using known angles from the unit circle. For instance, \(\cos \theta = \frac{\sqrt{2}}{2}\) yields \( \theta = \frac{\pi}{4} \) and \(\frac{7\pi}{4}\). Similarly, \(\cos \theta = -\frac{\sqrt{2}}{2}\) gives \(\theta = \frac{3\pi}{4}\) and \(\frac{5\pi}{4}\). Adding \(2k\pi\), where \(k\) is any integer, allows for expressing the generality of solutions, encompassing all angles that satisfy the original equation.