Problem 66 - Algebra 2

Question

a. Open-Ended Write three trigonometric equations whose complete solution is \(\pi+2 \pi n .\) b. Describe how you found the equations in part (a).

Step-by-Step Solution

Answer

A few examples of trigonometric equations whose complete solution is \( \pi + 2 \pi n \) are: 1) \( sin(x) = 0 \), 2) \( cos(x) = -1 \), and 3) \( 2sin(x) = 0 \). We found these equations by seeing that they have solutions that repeat every \(2 \pi\); they differ only by a shift (or phase) depending on the function and value we set the function equal to.

Step 1: Identifying the Trigonometric Functions

Think about trigonometric functions that have solutions every \(2 \pi\). The most basic examples are the sine (sin) and cosine (cos) functions. These functions have the same value every \(2 \pi\), so they will be a good starting point to form the equations. Consider the following three equations: 1) \(sin(x) = 0\) 2) \(cos(x) = -1\) 3) \(2sin(x) = 0\).

Step 2: Solving the Trigonometric Equations

Now, solve each of these equations to determine their complete solution. 1) \(sin(x) = 0\) has solutions \(x = 0, \pi, 2\pi, 3\pi, ..., n\pi\). When \(n\) is odd, the equation gives \(\pi\) and when \(n\) is even it gives \(0\). Hence, the solution could be expressed as \(x = \pi+2 \pi n\). 2) \(cos(x) = -1\) has solutions \(x = \pi, 3\pi, 5\pi, ..., (2n+1)\pi\). This can also be written as \(x = \pi+2 \pi n\). 3) \(2sin(x) = 0\) has the same solutions as the first equation since \(2sin(x) = 0\) reduces to \(sin(x) = 0\) and hence the solutions could also be expressed as \(x = \pi+2 \pi n\).

Step 3: Describing How the Equations Were Found

The equations were found by considering trigonometric functions that repeat every \(2 \pi\), specifically the sine and cosine functions. We manipulated these equations to have the form \(\pi+2 \pi n\) as the solution by setting the functions to specific values (0 and -1 in these cases). The '2' in the third equation doesn't change the solutions, and was added to show a variety of possible equations that satisfy the given form.

Key Concepts

sine function

The sine function, denoted as \( \sin(x) \), is one of the fundamental trigonometric functions that is periodic in nature. This means it repeats its values in regular intervals or cycles. Specifically, the sine function has a period of \(2\pi\), meaning after an interval of \(2\pi\), the function values repeat. The sine function takes an angle as input and returns a value ranging from -1 to 1. The graph of \( \sin(x) \) resembles a smooth wave that oscillates above and below the x-axis.

One key feature of \( \sin(x) \) is where it takes the value 0, which occurs at integer multiples of \( \pi \):

• \( \sin(0) = 0 \)
• \( \sin(\pi) = 0 \)
• \( \sin(2\pi) = 0 \)

These values are important when solving equations like \( \sin(x) = 0 \), as they form a pattern at intervals of \( \pi \). Therefore, the solutions to \( \sin(x) = 0 \) can be expressed succinctly as \( x = n\pi \), where \( n \) is an integer.

cosine function

The cosine function, denoted as \( \cos(x) \), is another fundamental trigonometric function which is also periodic with a period of \(2\pi\). The cosine function returns values ranging from -1 to 1, similar to the sine function, and its graph oscillates between these values in a smooth wave-like pattern.

A unique aspect of \( \cos(x) \) is where it reaches the value of -1, which occurs at the points:

• \( \cos(\pi) = -1 \)
• \( \cos(3\pi) = -1 \)
• \( \cos(5\pi) = -1 \)

These points demonstrate that the solutions to \( \cos(x) = -1 \) occur at \( x = (2n+1)\pi \), where \( n \) is an integer. This pattern translates to \( x = \pi + 2\pi n \), capturing the periodic repetition and providing an easy framework for solving cosine equations that evaluate to -1.

solutions of trigonometric equations

Solving trigonometric equations involves finding the angles that satisfy a given trig function equation. These solutions often utilize the periodic nature of trig functions such as sine and cosine, which repeat every \(2\pi\).

When solving an equation like \( \sin(x) = 0 \), we find that solutions occur at regular intervals along the x-axis (i.e., \( x = n\pi \)). This is due to the sine function crossing the x-axis at these points. Similarly, for \( \cos(x) = -1 \), solutions occur at intervals where the cosine function peaks at -1, like \( x = (2n+1)\pi \).

To generalize, one can express these solutions in the form \( x = \pi + 2\pi n \), indicating that every solution is a shift (or translation) of \( \pi \) added to the whole number multiple of the full cycle \(2\pi\). Understanding these patterns allows us to solve a wide variety of trigonometric equations efficiently by leveraging the inherent periodicity of the functions.