Problem 27 - Algebra 2

Question

Solve each equation for all values of \(\theta\). \(2 \cos ^{2} \theta+\cos \theta=0\)

Step-by-Step Solution

Answer

The solutions are \( \theta = \frac{\pi}{2} + n\pi \), \( \theta = \frac{2\pi}{3} + 2n\pi \), and \( \theta = \frac{4\pi}{3} + 2n\pi \), where \( n \) is an integer.

Step 1: Factor the Equation

First, we observe that the equation is a quadratic in terms of \( \cos \theta \). Therefore, we will factor this equation. Given: \( 2 \cos^2 \theta + \cos \theta = 0 \).We can factor out a \( \cos \theta \) from the equation:\( \cos \theta (2 \cos \theta + 1) = 0 \).

Step 2: Set Each Factor to Zero

Next, we solve each of the factors of the equation for \( \theta \):1. \( \cos \theta = 0 \)2. \( 2 \cos \theta + 1 = 0 \)

Step 3: Solve \( \cos \theta = 0 \)

The equation \( \cos \theta = 0 \) has solutions when \( \theta \) is an odd multiple of \( \frac{\pi}{2} \). Thus:\[ \theta = \frac{\pi}{2} + n\pi, \quad \text{where } n \text{ is an integer} \].

Step 4: Solve \( 2 \cos \theta + 1 = 0 \)

For the equation \( 2 \cos \theta + 1 = 0 \), solve for \( \cos \theta \):Subtract 1 from both sides: \( 2 \cos \theta = -1 \) Divide by 2: \( \cos \theta = -\frac{1}{2} \)This equation has solutions when \( \theta = \frac{2\pi}{3} + 2n\pi \) and \( \theta = \frac{4\pi}{3} + 2n\pi \), where \( n \) is an integer.

Key Concepts

Cosine Function

The cosine function, denoted as \( \cos \theta \), is one of the fundamental trigonometric functions used in mathematics. It relates the angle \( \theta \) of a right triangle to the ratio of the length of the adjacent side to the hypotenuse. Some key properties of cosine include:

• The cosine function is periodic with a period of \( 2\pi \). This means that its graph repeats every \( 2\pi \) units.
• It is defined for all real values of \( \theta \), but typically \( \theta \) is in radians for trigonometric applications.
• \( \cos(0) = 1 \) and \( \cos(\pi/2) = 0 \).
• The range of the cosine function is from \(-1\) to \(1\).

When solving trigonometric equations like \( \cos \theta = 0 \) or \( 2 \cos \theta + 1 = 0 \), understanding the cosine function's behavior helps us find the solutions for \( \theta \).

Factoring Quadratic Equations

Factoring quadratic equations is a key step in solving equations that can be expressed in the form \( ax^2 + bx + c = 0 \). When dealing with trigonometric equations, factors could include expressions like \( \cos \theta \).Here's a simple guide to factoring quadratic equations:

• Identify a common factor and factor it out. For example, in \( 2 \cos^2 \theta + \cos \theta = 0 \), \( \cos \theta \) is a common factor.
• Rewrite the equation as a product of factors. Here, it becomes \( \cos \theta (2 \cos \theta + 1) = 0 \).
• Use the Zero Product Property, which states if a product is zero, then at least one of its factors must be zero.

Factoring helps break down complex expressions into simpler ones, making it easier to apply solutions relevant to trigonometric identities and properties.

Solutions of Trigonometric Functions

Finding solutions to trigonometric equations involves identifying the values of \( \theta \) that satisfy equations like \( \cos \theta = 0 \) or \( \cos \theta = -\frac{1}{2} \).Here's how to solve them:

• For \( \cos \theta = 0 \), solutions are at odd multiples of \( \pi/2 \): \( \theta = \frac{\pi}{2} + n\pi \), where \( n \) is any integer.
• For \( \cos \theta = -\frac{1}{2} \), the solutions relate to specific angles like \( \frac{2\pi}{3} \) and \( \frac{4\pi}{3} \), extended by periods of \( 2\pi \): \( \theta = \frac{2\pi}{3} + 2n\pi \) and \( \theta = \frac{4\pi}{3} + 2n\pi \).

These solutions often rely on trigonometric identities and the periodic nature of functions like cosine, revealing multiple solutions across different cycles.

Phase Shift in Trigonometry

Phase shift is a horizontal translation of the graph of a trigonometric function.To understand phase shifts:

• In the standard cosine equation \( y = \cos(x) \), the graph is centered around the y-axis.
• A phase shift changes this position. If \( y = \cos(x + c) \), the graph shifts left by \( c \) units if \( c \) is positive, and right if \( c \) is negative.
• For example, if a solution involves \( \cos(\theta - \phi) \), then \( \phi \) is the phase shift.

Phase shifts help us model trigonometric equations more accurately by adjusting the 'start' of the cycle, especially useful in applications like signal processing or physics.