Problem 24 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find all solutions of the equation in the interval \([0,2 \pi)\). $$\sec x=\sqrt{2}$$

Step-by-Step Solution

Answer

The solutions of the equation \(\sec x=\sqrt{2}\) in the interval \([0, 2\pi)\) are \(x = \pi/4\) and \(x = 7\pi/4\).

Step 1: Convert secant into cosine

Since \(\sec x=\sqrt{2}\), we rewrite it in terms of cosine, obtaining \(\cos x = 1/\sqrt{2}\). This is done because cosine is a more familiar function and it is easier to solve for x.

Step 2: Solve for x

Now we must find all the values of x for which \(\cos x = 1/\sqrt{2}\). The solutions are \(x = \pi/4 + 2n\pi\) and \(x = 7\pi/4 + 2n\pi\), where n is an integer. However, regarding the prescribed interval of \([0, 2\pi)\), we only consider the solutions within this interval.

Step 3: Check the interval

After checking the interval, we eliminate the x-values that don't fall in the interval \([0, 2\pi)\). Therefore, \(x = \pi/4\) or \(x = 7\pi/4\) are the two points in the interval \([0, 2\pi)\) where \(\cos x = 1/\sqrt{2}\), thus they are the solutions to the initial problem. The secant is undefined where the cosine function equals zero, however none of our solutions have this issue.

Key Concepts

Secant Function

The secant function, denoted as \( \sec x \), is one of the trigonometric functions. In essence, the secant function is the reciprocal of the cosine function. This means that for any angle \( x \), \( \sec x = \frac{1}{\cos x} \). This relationship indicates that wherever the cosine function equals zero, the secant function will be undefined. In practical terms, understanding the secant function helps when you encounter trigonometric equations since it frequently requires conversion into a more recognizable function, like cosine.
For instance, when the exercise asked to solve \( \sec x = \sqrt{2} \), we recognized that working directly with the secant function isn't as straightforward as working with cosine. Therefore, by considering that \( \sec x = \frac{1}{\cos x} \), we convert the equation into \( \cos x = \frac{1}{\sqrt{2}} \). This makes the equation much simpler to solve, as it is expressed in terms of cosine. Converting between these functions is a common technique in trigonometry to simplify problem-solving.

Cosine Function

The cosine function is one of the principal trigonometric functions, often represented in the unit circle or in oscillations. It measures the horizontal distance, on a unit circle, from the origin to the circle's edge at a specific angle. The cosine function is vital because it allows us to find the cosine value for specific angles, either in degrees or radians (such as \( \cos x = \frac{1}{\sqrt{2}} \)).
When solving the given exercise about \( \sec x = \sqrt{2} \), the conversion to \( \cos x = \frac{1}{\sqrt{2}} \) is achieved by understanding the reciprocal relationship. This is more practical when you think of familiar angles that provide known cosine values, such as \( \frac{\pi}{4} \) and \( \frac{7\pi}{4} \), which equate to \( \frac{1}{\sqrt{2}} \). This method effectively plays on recognizing and making use of the cosine values from known angles, aiding in solving for \( x \) within a specified interval.

Interval Notation

Interval notation is a mathematical way to express a set of numbers lying within a given range. In this system, a pair of numbers is used to signify the lowest and highest number in the set. Closed intervals indicate that the endpoints are included in the set, denoted with square brackets (e.g., \([0, 2\pi)\)). Open intervals use parentheses, indicating the exclusion of boundaries. In trigonometric equations like ours, interval notation helps define the specific domain within which we seek solutions.
In the exercise, using the interval \([0, 2\pi)\), we specified that we want solutions only within these limits. This means \( x \) can be any value from 0 up to \( 2\pi \) (but not including \( 2\pi \) itself).
Such precise intervals ensure solutions are meaningful, preventing extraneous values that lie outside the desired range from confusing the outcome. This constraint is crucial, as trigonometric functions often have multiple solutions spanning wider domains.