Problem 110 - Contemporary Precalculus

Question

Let \(n\) be a fixed positive integer. Describe all solutions of the equation \(\sin n x=1 / 2 . \text { [Hint: See Exercises } 43-52 .]\)

Step-by-Step Solution

Answer

Answer: The general solution to the equation \(\sin nx = \frac{1}{2}\) is given by \(x=\frac{\frac{\pi}{6}+2\pi k_1}{n}\) or \(x=\frac{\frac{5\pi}{6}+2\pi k_2}{n}\), where \(k_1\) and \(k_2\) are integers.

Step 1: Recall sine function properties

The sine function takes values between -1 and 1. For an angle x, if \(\sin x = \frac{1}{2}\), then the possible values of x are given by \(x=\frac{\pi}{6}+2\pi k_1\) and \(x=\frac{5\pi}{6}+2\pi k_2\), where \(k_1\) and \(k_2\) are integers.

Step 2: Rewrite the equation in terms of angle x

The given equation is \(\sin nx = \frac{1}{2}\). We will replace the angle x in the properties mentioned in Step 1 with the angle nx.

Step 3: Find all possible solutions to the equation

According to the properties stated in step 1, the solutions of the given equation are \(nx =\frac{\pi}{6}+2\pi k_1\) and \(nx=\frac{5\pi}{6}+2\pi k_2\), where \(k_1\) and \(k_2\) are integers. Now, we will isolate x in both expressions: \(x = \frac{\frac{\pi}{6}+2\pi k_1}{n}\) and \(x=\frac{\frac{5\pi}{6}+2\pi k_2}{n}\).

Step 4: Write the general solution

Now we have the general solution to the given equation in terms of n and integers \(k_1\) and \(k_2\): \(x=\frac{\frac{\pi}{6}+2\pi k_1}{n}\) or \(x=\frac{\frac{5\pi}{6}+2\pi k_2}{n}\), where \(k_1\) and \(k_2\) are integers.

Key Concepts

Sine Function

The sine function, usually denoted as \(\sin\), is a fundamental part of trigonometry. It is defined for any real number and describes the ratio of the opposite side to the hypotenuse in a right triangle. The sine function takes values from -1 to 1 and has a periodic nature, repeating every \(2\pi\) radians. This periodicity means that for any angle \(x\), \(\sin(x) = \sin(x + 2\pi k)\), where \(k\) is an integer.

This property is crucial when solving equations involving the sine function because it enables us to find all the solutions to such equations by considering each full cycle of the sine wave. For example, if \(\sin x = \frac{1}{2}\), there will be multiple angles \(x\) that satisfy this equation, corresponding to different rotations of the circle. Understanding these properties makes it easier to manipulate and solve trigonometric equations.

One important point to remember is that the sine function has specific standard values at certain angles that are often used in trigonometry, like \(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \text{and} \frac{11\pi}{6}\), where the sine value is \(\frac{1}{2}\). Recognizing these angles helps quickly pinpoint solutions for more complex equations involving sine.

General Solutions

When solving trigonometric equations like \(\sin nx = \frac{1}{2}\), finding the general solution involves identifying all possible angles \(x\) that satisfy the equation. Given the periodic nature of the sine function, solutions come in families that are spaced regularly.

Let's examine the equation step by step. We start with noted values for \(\sin x = \frac{1}{2}\) at \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\). For the given equation \(\sin nx = \frac{1}{2}\), we substitute \(nx\) for \(x\), which gives us the angles \(nx = \frac{\pi}{6}\) and \(nx = \frac{5\pi}{6}\).

Because these solutions repeat every \(2\pi\), the general solution for \(nx\) is:

• \(nx = \frac{\pi}{6} + 2\pi k_1\)
• \(nx = \frac{5\pi}{6} + 2\pi k_2\)

By solving for \(x\) in these equations, we introduce integer variables \(k_1\) and \(k_2\) to account for every possible occurrence within the sine wave's cycles. Thus, we arrive at the general solution:
\[ x = \frac{\frac{\pi}{6} + 2\pi k_1}{n} \text{ or } x = \frac{\frac{5\pi}{6} + 2\pi k_2}{n} \]
Here, \(k_1\) and \(k_2\) cover every instance of the solution within all cycles of the sine function.

Angle Values

Understanding angle values is essential in solving trigonometric equations. Here, a specific focus is on angles related to the sine function that provides known results, like \(\sin x = \frac{1}{2}\). Standard angles such as \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\) are easily recognizable and have significance, as they regularly appear in sine function solutions.

When addressing equations such as \(\sin nx = \frac{1}{2}\), we are really dealing with variations of these standard angles. The equation manipulates these by using \(n\), and once we know the multiple of \(x\) that satisfies \(nx\), solving for \(x\) is a straightforward operation.

For example, when \(nx = \frac{\pi}{6}\), solving for \(x\) involves dividing both sides by \(n\) to achieve \(x = \frac{\frac{\pi}{6}}{n}\). The angle adjustments take into account the changes due to \(n\) and integer \(k\) values, covering all possible rotations within the trigonometric circle. These angles are fundamental, providing a roadmap to understanding and solving a wide range of trigonometric equations efficiently. Understanding these standard angle values is a great tool in your trigonometry toolkit.