Problem 31 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-38: Find all solutions of the equation. $$ (2 \sin \theta+1)(2 \cos \theta+3)=0 $$

Step-by-Step Solution

Answer

\(\theta = \frac{7\pi}{6} + 2n\pi\) and \(\theta = \frac{11\pi}{6} + 2n\pi\), where \(n\) is an integer.

Step 1: Understand the Problem

We need to find all the solutions for the equation \((2 \sin \theta + 1)(2 \cos \theta + 3) = 0\). This is a factored equation where the product of two expressions equals zero.

Step 2: Use the Zero Product Property

According to the zero product property, if a product of two factors is zero, at least one of the factors must be zero. Therefore, set each factor equal to zero: \(2 \sin \theta + 1 = 0\) and \(2 \cos \theta + 3 = 0\).

Step 3: Solve for \(\sin \theta\)

Solve the equation \(2 \sin \theta + 1 = 0\): 1. Subtract 1 from both sides: \ 2 \sin \theta = -1\. 2. Divide by 2: \ \sin \theta = -\frac{1}{2}\.

Step 4: Solve for \(\cos \theta\)

Solve the equation \(2 \cos \theta + 3 = 0\): 1. Subtract 3 from both sides: \ 2 \cos \theta = -3\. 2. Divide by 2: \ \cos \theta = -\frac{3}{2}\. However, note that the range for \(\cos \theta\) is \([-1,1]\), so there is no solution for this equation.

Step 5: Determine \(\theta\) Values for \(\sin \theta = -\frac{1}{2}\)

\(\sin \theta = -\frac{1}{2}\) corresponds to angles where the sine value is negative. The general solutions are: \ \theta = \frac{7\pi}{6} + 2n\pi, \text{ and } \theta = \frac{11\pi}{6} + 2n\pi\, where \(n\) is an integer.

Step 6: Compile the Solutions

The only valid solutions are derived from \(\sin \theta = -\frac{1}{2}\). Therefore, the solutions are: \ \theta = \frac{7\pi}{6} + 2n\pi \text{, and } \theta = \frac{11\pi}{6} + 2n\pi, \text{ where } n \text{ is an integer.}

Key Concepts

Zero Product Property

The Zero Product Property is a fundamental concept in algebra that plays a crucial role in solving equations like the one in this exercise. This property states that if the product of two expressions is zero, at least one of the expressions must be equal to zero. This is written mathematically as:

• If \(a \times b = 0\), then either \(a = 0\) or \(b = 0\) or both.

Using the Zero Product Property is particularly useful when solving factored equations, allowing us to split a complex equation into simpler parts. For the equation \((2 \sin \theta + 1)(2 \cos \theta + 3) = 0\), we apply this property to identify two separate equations:

• \(2 \sin \theta + 1 = 0\)
• \(2 \cos \theta + 3 = 0\)

Each factor is equated to zero, making it easier to solve for the unknown variable \(\theta\). This logical breakdown is what makes the Zero Product Property a powerful tool for solving equations effectively.

Sine Function

The Sine Function is one of the primary trigonometric functions, denoted as \(\sin(\theta)\), where \(\theta\) represents an angle. It is important for understanding periodic wave patterns, such as those found in sound waves or alternating current in electrical circuits. In the solution of the given exercise, we focused on solving the equation \(2 \sin \theta + 1 = 0\).Let's break it down:

• First, we isolate \(\sin \theta\) by subtracting 1 from both sides, resulting in \(2 \sin \theta = -1\).
• Next, divide by 2 to obtain \(\sin \theta = -\frac{1}{2}\).

The Sine Function has a range of \([-1, 1]\), so \(-\frac{1}{2}\) falls within this range, indicating valid solutions are possible. It's crucial to remember that the Sine Function is periodic, which means it repeats its values every \(2\pi\) radians. Therefore, solutions to the equation will occur at regular intervals, leading us to explore angle solutions in the next section.

Angle Solutions

Finding the Angle Solutions for trigonometric equations involves determining all possible angles that satisfy a given trigonometric expression. For our problem, we identified the expression \(\sin \theta = -\frac{1}{2}\). The goal is to find all angles \(\theta\) where this condition holds true, considering the periodic nature of trigonometric functions.For \(\sin \theta = -\frac{1}{2}\), the reference angle is \(\theta = \frac{\pi}{6}\). However, since the sine value is negative, \(\theta\) must be in the third or fourth quadrant:

• In the third quadrant, the angle is \( \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \).
• In the fourth quadrant, the angle is \( \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \).

Due to the periodicity of the sine function, these solutions repeat every full cycle of \(2\pi\), thus forming the general solutions:

• \(\theta = \frac{7\pi}{6} + 2n\pi\)
• \(\theta = \frac{11\pi}{6} + 2n\pi\)

where \(n\) is any integer. These solutions encompass all possible angles \(\theta\) that make \(\sin \theta = -\frac{1}{2}\) true, providing a comprehensive understanding of how angle solutions are determined for trigonometric equations.