Problem 14 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation in part (a) analytically over the interval \([0,2 \pi) .\) Then use a graph to solve each inequality in part (b). (a) \(\cos 2 x=-\frac{1}{2}\) (b) \(\cos 2 x>-\frac{1}{2}\)

Step-by-Step Solution

Answer

(a) Solutions: \(x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\). (b) Inequality solutions: \((0, \frac{\pi}{3}) \cup (\frac{2\pi}{3}, \frac{4\pi}{3}) \cup (\frac{5\pi}{3}, 2\pi)\).

Step 1: Understand the Equation

We need to solve the equation \(\cos 2x = -\frac{1}{2}\) for \(x\) in the interval \([0, 2\pi).\) This means finding all \(x\) such that when plugged into the cosine function, it results in \(-\frac{1}{2}.\)

Step 2: Find the General Solution for Cosine

The cosine function \(\cos(\theta)=-\frac{1}{2}\) has solutions at \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\). Because \(\cos(\theta)\) is periodic with period \(2\pi\), the general solution is \(\theta = \frac{2\pi}{3} + 2k\pi\) and \(\theta = \frac{4\pi}{3} + 2k\pi\) for \(k \in \mathbb{Z}.\)

Step 3: Solve for 2x

We set \(2x = \frac{2\pi}{3} + 2k\pi\) and \(2x = \frac{4\pi}{3} + 2k\pi\). Dividing by 2, we have:\[x = \frac{\pi}{3} + k\pi\quad \text{and} \quad x = \frac{2\pi}{3} + k\pi.\]

Step 4: Find Specific Solutions in Given Interval

Within the interval \([0, 2\pi)\), substitute different values for \(k\) to find solutions:- For \(x = \frac{\pi}{3} + k\pi\): - \(k=0\), \(x = \frac{\pi}{3}\). - \(k=1\), \(x = \frac{4\pi}{3}\).- For \(x = \frac{2\pi}{3} + k\pi\): - \(k=0\), \(x = \frac{2\pi}{3}\). - \(k=1\), \(x = \frac{5\pi}{3}\).The solutions are therefore \(x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}.\)

Step 5: Use the Graph for Inequality

For the inequality \(\cos 2x > -\frac{1}{2}\), we consider the graph of \(y = \cos 2x\). The inequality holds where the graph is above the horizontal line \(y = -\frac{1}{2}\). This occurs between the values of \(x\) found in Step 4 and also where the cosine function remains above \(-\frac{1}{2}\) due to its periodic nature.

Step 6: Determine Interval for Inequality

The cosine \(2x > -\frac{1}{2}\) is true between each pair of consecutive solutions from part (a). Hence, the solution set for the inequality is:\(\left(0, \frac{\pi}{3}\right) \cup \left(\frac{2\pi}{3}, \frac{4\pi}{3}\right) \cup \left(\frac{5\pi}{3}, 2\pi\right).\)

Key Concepts

Cosine Function

The cosine function, represented as \(\cos(\theta)\), is a fundamental trigonometric function that maps angles to a value between -1 and 1. It is an even function, meaning that \(\cos(-\theta) = \cos(\theta)\). This symmetry is around the vertical axis, making it an important property when solving trigonometric equations. The graph of the cosine function is a wave that repeats every \(2\pi\) radians. This makes it ideal for describing periodic phenomena, such as sound waves and seasonal patterns.

When solving equations like \(\cos(2x) = -\frac{1}{2}\), we use properties of the cosine function. Here, we specifically look for angles where the cosine value is \(-\frac{1}{2}\), and this can help us find all possible solutions within a given range. The cosine of \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\) are both \(-\frac{1}{2}\), illustrating solutions at these key angles within a complete cycle.

General Solution

In trigonometry, finding a general solution to an equation involves understanding the periodic nature of trigonometric functions. For instance, the cosine function is periodic with a period of \(2\pi\), meaning every \(2\pi\) units, the function repeats its values. This is critical when solving an equation like \(\cos(\theta) = -\frac{1}{2}\).

For \(\theta\), where \(\cos(\theta) \) is \(-\frac{1}{2}\), the general solution can be written as:

• \(\theta = \frac{2\pi}{3} + 2k\pi\)
• \(\theta = \frac{4\pi}{3} + 2k\pi\)

where \(k\) is an integer. This represents all the angles that will satisfy the original condition by accounting for the whole infinite set of angle solutions due to the periodic nature.

Periodic Functions

Periodic functions, like the cosine function, repeat their values over regular intervals, known as periods. The period of the cosine function is \(2\pi\) radians, meaning its values repeat every \(2\pi\) change in the angle. This repetitive pattern is essential in solving trigonometric equations, especially when we want to find solutions over a specific interval.

Understanding this concept helps us realize that if we have a solution for a trigonometric equation, such as \(x = \frac{\pi}{3}\), then \(x + 2k\pi\) will also be solutions for any integer \(k\). This periodic property thus enables us to fill in all anti-nodes of the trigonometric wave for the entire domain by understanding just one cycle.

Interval Notation

Interval notation is a simplified way of representing continuous sets of numbers and is especially useful in describing the solutions to inequalities. In this context, after solving \(\cos 2x > -\frac{1}{2}\) graphically, we express our solutions using interval notation to concisely list the valid ranges of \(x\).

When the graph of \(y = \cos 2x\) is seen above the line \(y = -\frac{1}{2}\), we determine intervals such as:

• \((0, \frac{\pi}{3})\)
• \((\frac{2\pi}{3}, \frac{4\pi}{3})\)
• \((\frac{5\pi}{3}, 2\pi)\)

These intervals specify where the inequality is true, using round brackets to indicate that endpoints are not included. Interval notation provides a precise, concise way to express these ranges for anyone solving similar equations.