Problem 19 - Precalculus Mathematics for Calculus

Question

An equation is given. (a) Find all solutions of the equation. (b) Find the solutions in the interval \([0,2 \pi)\) $$2 \cos 2 \theta+1=0$$

Step-by-Step Solution

Answer

The solutions are \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \text{and} \frac{5\pi}{3}\).

Step 1: Simplify the Equation

The given equation is \(2 \cos 2\theta + 1 = 0\). To isolate the trigonometric function, subtract 1 from both sides: \(2 \cos 2\theta = -1\). Then, divide both sides by 2: \(\cos 2\theta = -\frac{1}{2}\).

Step 2: Find General Solutions for \(2\theta\)

The equation \(\cos 2\theta = -\frac{1}{2}\) is a standard cosine equation. The solutions to \(\cos x = -\frac{1}{2}\) are given by \(x = \frac{2\pi}{3} + 2k\pi\) and \(x = \frac{4\pi}{3} + 2k\pi\) for integers \(k\). Therefore, \(2\theta = \frac{2\pi}{3} + 2k\pi\) or \(2\theta = \frac{4\pi}{3} + 2k\pi\).

Step 3: Solve for \(\theta\)

To find \(\theta\), divide all parts of the general solutions by 2: \(\theta = \frac{\pi}{3} + k\pi\) and \(\theta = \frac{2\pi}{3} + k\pi\). These represent the general solutions for \(\theta\).

Step 4: Determine Solutions in the Interval \([0, 2\pi)\)

We need to identify which solutions satisfy the interval \([0, 2\pi)\). Consider the integer values of \(k\) for both solutions: - For \(\theta = \frac{\pi}{3} + k\pi\), the valid intervals occur at \(k=0\) \((\theta = \frac{\pi}{3})\) and \(k=1\) \((\theta = \frac{4\pi}{3})\).- For \(\theta = \frac{2\pi}{3} + k\pi\), the valid intervals occur at \(k=0\) \((\theta = \frac{2\pi}{3})\) and \(k=1\) \((\theta = \frac{5\pi}{3})\).Thus, the solutions in the interval \([0, 2\pi)\) are \(\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \text{and} \frac{5\pi}{3}\).

Key Concepts

Cosine Function

The cosine function is one of the fundamental trigonometric functions and is often used to describe the relationship between angles and side lengths in right-angled triangles. It is defined as the ratio of the adjacent side to the hypotenuse. In the unit circle, the cosine of an angle

• is the x-coordinate of a point lying on the circle's circumference.
• varies between -1 and 1 for angles from 0 to 360 degrees or from 0 to \(2\pi\) radians.

When it comes to equations like \(2 \cos 2\theta + 1 = 0\), the aim is to solve for the angle \(\theta\) that makes the cosine expression a specific value. Initially, the equation is simplified to find \(\cos 2\theta = -\frac{1}{2}\), which means finding angles whose cosine is \(-\frac{1}{2}\). This involves understanding the behavior of the cosine function across different angles, as the function periodically repeats every \(2\pi\) radians (or 360 degrees). Thus, understanding the properties of the cosine function is critical for analyzing and solving trigonometric equations.

General Solutions

Finding general solutions for trigonometric equations involves identifying all possible values that satisfy the equation, often expressed in terms of a variable that represents the periodic nature of trigonometric functions.
In the case of the equation \(\cos 2\theta = -\frac{1}{2}\), once simplified, it reflects a standard angle problem that occurs at specific known positions, such as \(x = \frac{2\pi}{3}\) and \(x = \frac{4\pi}{3}\).
This is because the cosine of these angles gives exactly \(-\frac{1}{2}\).

To account for the infinite nature of angles, solutions are expressed using the integer variable, \(k\), reflecting periods of repetition:

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

These solutions show how every possible angle can be generated by adding full rotation periods \(2\pi\) to the base angles. By dividing these solutions by 2, we obtain the \(\theta\) values, which are the true general solutions: \(\theta = \frac{\pi}{3} + k\pi\) and \(\theta = \frac{2\pi}{3} + k\pi\). These generic solutions represent all instances where the cosine equals \(-\frac{1}{2}\), for all \(k\).

Interval Solutions

Interval solutions focus on finding specific angles or values of \(\theta\) that satisfy the initial trigonometric equation within a set period or range. In this instance, we are interested in solutions that fall within the interval \([0, 2\pi)\).

The previously calculated general solutions for \(\theta\):

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

These encompass infinite solutions, spanning every \(k\) value. However, focusing on specific \(k\) ensures solutions are within \([0, 2\pi)\).For \(\theta = \frac{\pi}{3} + k\pi\):

• \(k=0\) gives \(\theta = \frac{\pi}{3}\).
• \(k=1\) gives \(\theta = \frac{4\pi}{3}\).

For \(\theta = \frac{2\pi}{3} + k\pi\):

• \(k=0\) provides \(\theta = \frac{2\pi}{3}\).
• \(k=1\) provides \(\theta = \frac{5\pi}{3}\).

Thus, the specific solutions within the desired interval \([0, 2\pi)\) are \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3},\) and \(\frac{5\pi}{3}\). This narrows down the infinite possibilities to feasible angles for practical application, meeting the equation's requirements within the specified boundary.