Problem 54 - Precalculus: A Concise Course

Question

Use a graph to solve the equation on the interval \([-2 \pi, 2 \pi]\). $$ \sec x=2 $$

Step-by-Step Solution

Answer

The solutions to the equation \(\sec(x) = 2\) in the interval \([-2 \pi, 2 \pi]\) are \(x = -5 \pi / 3, \pi / 3, 5 \pi / 3\).

Step 1: Transform the Equation

Transform the given equation by expressing the secant function in terms of the cosine function. The secant function is the reciprocal of the cosine function, so the equation \(\sec x = 2\) becomes \(1 / cos(x) = 2\).

Step 2: Solve for Cosine

Solve the equation for cosine, which yields \(cos(x) = 1 / 2\). Cosine equals 1/2 at x values of \(\pi / 3\) and \(5 \pi / 3\) in the domain of \([-\pi, \pi]\).

Step 3: Extend the Solution for the Given Interval

Extend the solution for the given interval. Because cosine is a periodic function with a period of \(2 \pi\), if \(cos(x1) = cos(x2)\), then \(x2 = x1 + 2 \pi k\), where k is any integer. Thus, the full solutions in the interval \([-2 \pi, 2 \pi]\) would be \(x = \pi / 3 + 2 \pi k\) and \(x = 5 \pi / 3 + 2 \pi k\), where \(k = -1, 0, 1\).

Step 4: Plot the Solution

Plot the function \(\sec x\) or \(1 / cos(x)\) and the line \(y = 2\). The intersection points are the solutions. In the interval \([-2 \pi, 2 \pi]\), there are three intersection points, namely \(-5 \pi / 3, \pi / 3, 5 \pi / 3\). This confirms the solutions found in previous steps.

Key Concepts

Secant Function

The secant function, denoted as \( \sec(x) \), is closely related to the cosine function. It represents the reciprocal of the cosine of an angle in a right-angled triangle or on the unit circle. That means \( \sec(x) = \frac{1}{\cos(x)} \).

• When the cosine value is large, the secant value is small, and vice versa.
• Secant is undefined for angles where cosine is zero, because division by zero is undefined.
• It has vertical asymptotes at these points, which occur at odd multiples of \( \pi/2 \).

To solve problems involving secant functions, we often convert them to cosine functions since cosine is often easier to interpret and calculate. This conversion is why in our exercise \( \sec(x)=2 \) became \( \cos(x)=\frac{1}{2} \).

Cosine Function

The cosine function, \( \cos(x) \), is one of the fundamental trigonometric functions. It describes the horizontal coordinate of a point on the unit circle corresponding to an angle \( x \).

• Cosine is periodic with a period of \( 2\pi \), which means \( \cos(x) = \cos(x + 2\pi k) \) for any integer \( k \).
• \( \cos(x) \) ranges from -1 to 1.
• The cosine function is symmetric about the y-axis (an even function), so \( \cos(-x) = \cos(x) \).

In solving \( \cos(x)=\frac{1}{2} \), we know from trigonometric values that \( \cos(\pi/3) = \frac{1}{2} \). Multiple equivalent angles within the given interval can be found by considering this symmetry and periodicity.

Interval Analysis

In trigonometric problems, interval analysis involves finding all possible solutions within a specified range. The exercise examines the interval \([-2\pi, 2\pi]\), which spans over more than one full circle of the unit circle.

• This interval includes negative and positive angles, covering both clockwise and counterclockwise views.
• Because trigonometric functions like cosine are periodic, solutions will repeat every \( 2\pi \).

For our exercise, once we found the primary solutions \( x = \pi/3 \) and \( x = 5\pi/3 \) within the primary cycle, we needed to extend these solutions across the specified interval.By adding multiples of \( 2\pi \) (positive and negative), additional solutions were found within the interval, leading to solutions like \(-5\pi/3\), ensuring all possible angles in the range where \( \sec(x) = 2 \) are accounted for.

Graphical Solutions

Finding graphical solutions involves visualizing equations to understand where they intersect, yielding possible solutions. In a graph, points where \( \sec(x) \) or \( \frac{1}{\cos(x)} \) equals a constant, such as 2, are shown as intersection points.

• Plotting \( \sec(x) \) results in a series of waves with vertical asymptotes at points where cosine is zero.
• The line \( y = 2 \) is straight and horizontal.
• The points where this line intersects the curve of \( \sec(x) \) give the solution to the equation \( \sec x = 2 \).

In our exercise, graphing these functions visually confirms the algebraic solutions \(-5\pi/3\), \(\pi/3\), and \(5\pi/3\), showing where \( \sec \) reaches a value of 2 on the interval \([-2\pi, 2\pi]\). This method can be particularly helpful for relating algebraic solutions to visual insights.