Problem 25 - BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Question

What are the solutions of the equation \(2 \sin x-1=0\) for \(0 \leq x<2 \pi\) ? (A) \(\frac{\pi}{3}\) (B) \(\frac{\pi}{6}\) (C) \(\frac{2 \pi}{3}\) (D) \(\frac{5 \pi}{6}\)

Step-by-Step Solution

Answer

The solutions of the equation \(2 \sin x-1=0\) for \(0 \leq x < 2 \pi\) are \(x = \frac{\pi}{6}\) and \(x = \frac{5 \pi}{6}\). So the correct answers are options (B) and (D).

Step 1: Set up the equation

Start by setting the equation \(2 \sin x-1=0\) equal to each other. Adding 1 to both sides we get \(2 \sin x = 1\). Now divide both sides by 2 to isolate and solve for \(\sin x\). This leads us to \(\sin x = \frac{1}{2}\).

Step 2: Find the angles which the sine function equates to 1/2

Now we look for the values \(x\) in the interval \(0 \leq x < 2 \pi\), for which the sine function equals 1/2. We know from the unit circle that \(\sin x = \frac{1}{2}\) at \(x = \frac{\pi}{6}\) and \(x = \frac{5 \pi}{6}\) within the given interval.

Step 3: Double-check all solutions and answer the question

Once the potential solutions have been identified, it is necessary to confirm they are indeed true solutions and also to confirm whether they fall within the specified domain. In this case, both the solutions, \(x = \frac{\pi}{6}\) and \(x = \frac{5 \pi}{6}\), are within the range \(0 \leq x < 2 \pi\). The solutions are given by options (B) and (D).

Key Concepts

Unit Circle

The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane. It is an essential tool in trigonometry for understanding the relationships between angles and the values of sine, cosine, and tangent functions.
The x-coordinate of a point on the unit circle represents the cosine of the angle, while the y-coordinate represents the sine of the angle.
For any angle \( x \) on this circle, the length of the radius is 1, simplifying calculations.

• The whole circle represents an angle from 0 to \( 2\pi \).
• Key angles used in trigonometry often include \( \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{4}, \) and so on.

Understanding the unit circle is foundational for solving trigonometric equations since it provides a visual way to see where certain values occur.

Sine Function

The sine function is a crucial concept in trigonometry and is defined as the vertical coordinate (y-value) of a point on the unit circle.
Given an angle \( x \), \( \sin x \) gives you the height of the point above the x-axis.

• The sine function has a range of [-1, 1], indicating how high or low the point can be on the unit circle.
• It is periodic with a period of \( 2\pi \), meaning that after every \( 2\pi \) units, the function's values repeat.
• Key values to remember are \( \sin \frac{\pi}{6} = \frac{1}{2} \) and \( \sin \frac{5\pi}{6} = \frac{1}{2} \).

These points help in identifying solutions for angles whose sine equals \( \frac{1}{2} \). Understanding these concepts allows for solving trigonometric equations systematically.

Solutions of Equations

Finding solutions to trigonometric equations like \( 2 \sin x - 1 = 0 \) involves a step-by-step process. First, we aim to isolate the trigonometric function.
By solving \( 2 \sin x = 1 \), we divide both sides by 2 to get \( \sin x = \frac{1}{2} \).

• Next, use your knowledge of the unit circle and sine function to find angles where \( \sin x \) equals these values.
• In this case, the unit circle tells us that these are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).
• Verify these angles fall within the given interval \( 0 \leq x < 2\pi \).

Revisiting the interval makes sure that no potential solutions are missed or mistakenly included.

Interval Notation

Interval notation is a way to represent a set of numbers, often corresponding to solutions of equations or valid domains.
This notation is important for clearly communicating the range in which solutions exist.

• Square brackets [ ] are used to include the endpoint in the interval (closed interval).
• Round brackets ( ) are used to exclude the endpoint (open interval).
• In the equation, the solution is \( 0 \leq x < 2\pi \), which means \( x \) starts at 0 and is less than, but not equal to, \( 2\pi \).

Using interval notation helps in precisely defining the solution set, ensuring all possible solutions are correctly identified and interpreted.