Problem 24 - Algebra 2

Question

Solve each equation for \(0 \leq \theta<2 \pi\). $$ 2 \sin \theta=3 $$

Step-by-Step Solution

Answer

The equation \(2 \sin \theta = 3\) has no solution for real values of \(\theta\), because the sine of a real number always falls within the range \[-1, 1\], while here it is required to equal \(\frac{3}{2}\).

Step 1: Isolate the Trigonometric Function

Rearrange the given equation to isolate \(\sin \theta\) on one side: \(\sin \theta = \frac{3}{2}\)

Step 2: Calculate Inverse of Sine

For real-valued \(\theta\), \(\sin \theta\) lies within the range \[-1, 1\]. However, here \(\sin \theta = \frac{3}{2}\) is outside this range, thus it has no solution.

Key Concepts

Circular Functions

Circular functions, often known as trigonometric functions, describe the relationships of angles and sides within a circle. They include sine, cosine, and tangent, among others. These functions are foundational in understanding how angles work in the unit circle.
Trigonometric equations are often solved within a specific interval, typically ranging from 0 to 2π, to find the true angles that satisfy certain conditions. This range corresponds to one full rotation around the circle. Circular functions repeat their values in a predictable cycle called a periodic or circular interval, making them particularly useful in understanding concepts that involve cyclical patterns.
When working with circular functions, you will often be tasked with converting between degrees and radians since radians are mathematically cleaner and more elegant in higher mathematics. Remember:

• 1 full circle is 360° or 2π radians.
• 90° is equivalent to π/2 radians.
• 180° is equivalent to π radians.

Using circular functions in problem-solving involves identifying the correct intervals and employing them to find relevant angles that satisfy specific equations.

Algebra 2

In Algebra 2, students deepen their understanding of algebraic principles, equipping them with tools to tackle complex equations, including those involving trigonometric functions. A standard exercise might require students to isolate variables, manipulate equations, and understand functional ranges.
Consider the original problem: solving the equation \[2 \sin \theta = 3\]To find solutions, you rearrange it to \[\sin \theta = \frac{3}{2}\]In Algebra 2, students become familiar with not just numerical solutions, but conditions where solutions may exist or not based on mathematical principles. This relates to knowing the typical range of sine functions, which is limited to \[[-1, 1]\] for real numbers. Algebra 2 encourages logical thinking and problem-solving skills—essentials in evaluating when an equation has or does not have legitimate solutions.

Sine Function

The sine function is one of the primary trigonometric functions used to relate an angle of a right triangle to the ratio of the length of the opposite side to the hypotenuse. It is represented as \[\sin \theta\]Understanding this function requires knowing its range, periodicity, and its behavior within the unit circle. The standard range of sine is from -1 to 1, meaning any value outside this interval cannot be the sine of a real number.

• At 0° or 0 radians, \[\sin \theta = 0\]
• At 90° or π/2 radians, \[\sin \theta = 1\]
• The sine function is periodic with a period of 2π, completing a full cycle every 360° or 2π radians.

In our exercise, when asked to solve \[\sin \theta = \frac{3}{2}\], it’s essential to recognize that \[\frac{3}{2}\] is beyond the maximum possible value of 1 for sine, indicating no solutions exist for real values of \(\theta\). This demonstrates the importance of understanding the limitations of trigonometric functions when solving equations.