Problem 22 - Algebra and Trigonometry

Question

\(17-34\) . An equation is given. (a) Find all solutions of the equation. (b) Find the solutions in the interval \([0,2 \pi) .\) $$ \sec 4 \theta-2=0 $$

Step-by-Step Solution

Answer

Solutions are \(\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}\).

Step 1: Rearrange the Equation

Start by rewriting the given equation: \( \sec 4\theta - 2 = 0 \). Add 2 to both sides to isolate the secant term: \( \sec 4\theta = 2 \).

Step 2: Convert Secant to Cosine

Recall that \( \sec \theta = \frac{1}{\cos \theta} \). Therefore, \( \sec 4\theta = 2 \) can be rewritten as \( \frac{1}{\cos 4\theta} = 2 \), which simplifies to \( \cos 4\theta = \frac{1}{2} \).

Step 3: Solve for Angle Function

We need to find when \( \cos 4\theta = \frac{1}{2} \). The cosine function equals \( \frac{1}{2} \) at angles \(\theta = \frac{\pi}{3} + 2k\pi \) and \(\theta = -\frac{\pi}{3} + 2k\pi\) for any integer \(k\). Therefore, \(4\theta = \frac{\pi}{3} + 2k\pi\) and \(4\theta = -\frac{\pi}{3} + 2k\pi\).

Step 4: Solve for \( \theta \)

Divide both sides of each equation by 4 to solve for \( \theta \):1. \( \theta = \frac{\pi}{12} + \frac{k\pi}{2} \) 2. \( \theta = -\frac{\pi}{12} + \frac{k\pi}{2} \)

Step 5: Find Solutions within the Interval \([0, 2\pi)\)

We need solutions for \( \theta \) in the interval \([0, 2\pi)\). Substitute integer values for \(k\) and check each:1. For \( \theta = \frac{\pi}{12} + \frac{k\pi}{2} \): - \(k = 0: \theta = \frac{\pi}{12} \) - \(k = 1: \theta = \frac{7\pi}{12} \) - \(k = 2: \theta = \frac{13\pi}{12} \) - \(k = 3: \theta = \frac{19\pi}{12} \)2. For \( \theta = -\frac{\pi}{12} + \frac{k\pi}{2} \): - \(k = 1: \theta = \frac{5\pi}{12} \) - \(k = 2: \theta = \frac{11\pi}{12} \) - \(k = 3: \theta = \frac{17\pi}{12} \)

Key Concepts

Secant Function

The secant function, written as \( \sec \theta \), is a fundamental concept in trigonometry. It is defined as the reciprocal of the cosine function. In simpler terms, \( \sec \theta = \frac{1}{\cos \theta} \). This definition tells us that wherever the cosine function is zero, the secant function is undefined, as division by zero is not allowed in mathematics.

• The secant function corresponds to the hypotenuse over the adjacent side of a right triangle when considering \( \theta \) as an angle.
• The graph of the secant function shows distinct features such as vertical asymptotes where cosine equals zero.

Understanding the relationship between secant and cosine is crucial when solving trigonometric equations as it allows us to convert between these two forms, facilitating the solution process. In the given exercise, converting \( \sec 4\theta = 2 \) to its cosine equivalent \( \cos 4\theta = \frac{1}{2} \) simplifies solving the problem.

Cosine Function

The cosine function, represented by \( \cos \theta \), is one of the primary trigonometric functions used in mathematics. This function gives the ratio of the adjacent side of a right triangle to its hypotenuse, associated with an angle \( \theta \).

• The cosine function is periodic, meaning it repeats its values in a regular pattern.
• It has a period of \( 2\pi \), which means every \( 2\pi \) units, the cosine values repeat.

In the context of solving equations, finding specific angles where the cosine is a known value is a common task. For example, \( \cos 4\theta = \frac{1}{2} \) requires identifying standard angles from unit circle conventions. These angles include \( \theta = \frac{\pi}{3} \) and \( \theta = \frac{5\pi}{3} \) in one period. After solving for these angles, general solutions introduce terms like \( 2k\pi \) to incorporate all possible angles fitting this cosine value.

Interval Notation

Interval notation is a mathematical way of expressing a range of values along the number line, particularly valuable when specifying domains or solutions within certain bounds. In the exercise, we need to identify specific solutions within the interval \([0, 2\pi)\).

• Brackets \([\ ]\) indicate that the endpoint values are included in the interval.
• Parentheses \((\ )\) indicate that the endpoint values are not included.

For example, \([0, 2\pi)\) means all values starting from 0 up to but not including \(2\pi\). When solving trigonometric equations, it's crucial to check that our solutions fall within such specified intervals. In this case, verifying solutions like \( \theta = \frac{\pi}{12} \) and \( \theta = \frac{7\pi}{12} \) ensures they are valid within the given limits, guiding us to provide accurate and meaningful solutions.