Problem 104 - Precalculus with Limits

Question

Consider the equation \( 2 \sin x - 1 = 0 \). Explain the similarities and differences between finding all solutions in the interval \( \left[0,\dfrac{\pi}{2}\right) \), finding all solutions in the interval \( [0,2\pi) \), and finding the general solution.

Step-by-Step Solution

Answer

The solutions of the equation \(2 \sin x - 1 = 0\) are \(x = \frac{\pi}{6}\) in the interval [0, \(\frac{\pi}{2}\)], \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\) in the interval [0, 2\pi), and x = n\(2\pi\) + \(\frac{\pi}{6}\) and x = n\(2\pi\) + \(\frac{5\pi}{6}\) as the general solution, where n is an integer.

Step 1: Solving \(2 \sin x - 1 = 0\) in [0, \(\frac{\pi}{2}\)]

To solve for x in the first interval, first, isolate \(\sin x\) by adding 1 to both sides of the equation and then dividing by 2: \(\sin x = \frac{1}{2}\). Use the unit circle or sine function properties to determine that the solution in the interval [0, \(\frac{\pi}{2}\)] is \(x = \frac{\pi}{6}\). In this interval, the sine function starts at 0, increases to 1 at \(\frac{\pi}{2}\), during which it passes \(\frac{1}{2}\) once.

Step 2: Solving \(2 \sin x - 1 = 0\) in [0, 2\pi)

In the interval [0, 2\pi), the sine function completes a whole cycle, starting and ending at 0, increasing to 1, then decreasing to -1, and back to 0. So, it passes \(\frac{1}{2}\) twice - once while increasing, and the other while decreasing. Therefore, there are two solutions in this interval, which are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).

Step 3: Finding the General Solution for \(2 \sin x - 1 = 0\)

The general solution contains all possible solutions in the form x = n\(2\pi\) + \(\frac{\pi}{6}\), and x = n\(2\pi\) + \(\frac{5\pi}{6}\), where n is an integer. This shows the repeated or cyclic nature of the sine function for every \(2\pi\) interval.

Key Concepts

Unit Circle

The unit circle is a fundamental concept in trigonometry. It is a circle with a radius of 1 and is centered at the origin of a coordinate plane. This circle helps us visualize angles and their trigonometric ratios.
For the unit circle:

• The x-coordinate represents the cosine of an angle.
• The y-coordinate represents the sine of an angle.

This makes it easy to find the sine or cosine of standard angles, like those commonly used in exercises.

When solving the equation in [0, \(2\pi)\), the sine function repeats its values, completing a full wave cycle as it traces out the unit circle. For \(\sin x = \frac{1}{2}\), you identify the corresponding points where the sine value is reached. The first is at \(\frac{\pi}{6}\) or 30 degrees, and the second is at \(\frac{5\pi}{6}\) or 150 degrees. These points lie at intersections on the unit circle where the y-axis value matches \(\frac{1}{2}\).

Sine Function

The sine function is a periodic function that describes the y-coordinate on the unit circle. It varies smoothly between -1 and 1.

• The sine of 0 is 0.
• The maximum value of 1 occurs at \(\pi/2\).
• The minimum value of -1 occurs at \(3\pi/2\).

In the interval \([0, \frac{\pi}{2})\), the sine function increases from 0 to 1. At this stage, the equation \(2\sin x - 1 = 0\) simplifies to \(\sin x = \frac{1}{2}\). The only solution here is \(x = \frac{\pi}{6}\).

In a complete cycle \([0, 2\pi)\), the sine curve hits \(\frac{1}{2}\) twice: once as it ascends to 1, and then again as it descends after peaking. These crossings correspond to \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).

General Solution

The general solution to the trigonometric equation involves finding all possible values for which \(\sin x = \frac{1}{2}\) holds true repeatedly in cycles.
The sine function is periodic with a period of \(2\pi\). Thus, any angle that solves the equation will repeat after every \(2\pi\) intervals. So, the general solutions are given by:

• \(x = n2\pi + \frac{\pi}{6}\)
• \(x = n2\pi + \frac{5\pi}{6}\)

Here, \(n\) is any integer that shifts the angles by complete cycles of \(2\pi\).
This reflects the cyclic nature of trigonometric functions, allowing it to provide solutions infinitely for any integer \(n\). This is essential for defining solutions beyond any finite interval, covering all possible occurrences of the sine value \(\frac{1}{2}\).