Problem 2 - Precalculus

Question

In Exercises \(1-18,\) find all of the exact solutions of the equation and then list those solutions which are in the interval \([0,2 \pi)\). $$ \cos (3 x)=\frac{1}{2} $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}, \frac{11\pi}{9}, \frac{13\pi}{9}, \frac{17\pi}{9}\).

Step 1: Understand the Cosine Equation

The given equation is \( \cos(3x) = \frac{1}{2} \). We need to find all solutions for \(x\) that satisfy this equation and also lie within the interval \([0, 2\pi)\).

Step 2: Solve for 3x Using the General Cosine Solution

Recall that \( \cos(\theta) = \frac{1}{2} \) is true for \( \theta = \frac{\pi}{3} + 2k\pi \) and \( \theta = -\frac{\pi}{3} + 2k\pi \), where \(k\) is any integer. Therefore, for \( 3x \), we write:\[ 3x = \frac{\pi}{3} + 2k\pi \] \[ 3x = -\frac{\pi}{3} + 2k\pi \].

Step 3: Solve for x

Divide each part of the equations from Step 2 by 3 to solve for \(x\):\[ x = \frac{\pi}{9} + \frac{2k\pi}{3} \] \[ x = -\frac{\pi}{9} + \frac{2k\pi}{3} \].

Step 4: Find Solutions in the Interval [0, 2π)

Substitute different integer values of \(k\) into both equations from Step 3 and find values of \(x\) that lie within the interval \([0, 2\pi)\).- For \( k = 0 \): \[ x = \frac{\pi}{9} \ and \ x = -\frac{\pi}{9} + \frac{2\pi}{3} = \frac{5\pi}{9} \]- For \( k = 1 \): \[ x = \frac{\pi}{9} + \frac{2\pi}{3} = \frac{7\pi}{9} \ and \ x = \frac{5\pi}{9} + \frac{2\pi}{3} = \frac{11\pi}{9} \]- For \( k = 2 \): \[ x = \frac{\pi}{9} + \frac{4\pi}{3} = \frac{13\pi}{9} \ and \ x = \frac{5\pi}{9} + \frac{4\pi}{3} = \frac{17\pi}{9} \]The solutions \( x = \frac{19\pi}{9}, \frac{23\pi}{9}, \frac{25\pi}{9} \) for \( k = 3, 4, ... \) exceed \(2\pi\) and are not within the interval.

Step 5: List All Valid Solutions

Collect the solutions that lie within the interval \([0, 2\pi)\):- \(x = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}, \frac{11\pi}{9}, \frac{13\pi}{9}, \frac{17\pi}{9} \).

Key Concepts

Cosine Function

The cosine function, often denoted as \( \cos \), is one of the primary trigonometric functions. It measures the adjacent side over the hypotenuse in a right-angled triangle or relates the coordinates of a point on the unit circle.
The cosine function has a few key properties:

• It is periodic with a period of \( 2\pi \).
• The range of the cosine function is \([-1, 1]\).
• The cosine of an angle \( \theta \) gives the x-coordinate of a point on the unit circle.

Understanding these properties helps us solve trigonometric equations by recognizing where specific values, like \( \frac{1}{2} \), occur within one cycle of the function.
In our exercise, being familiar with the fact that \( \cos(\theta) = \frac{1}{2} \) occurs at specific angles helps identify the general solutions.

General Solution

When solving trigonometric equations, finding the general solution is key. For the cosine function, the general solution refers to all the possible angles that satisfy the given trigonometric equation, expressed in terms of a variable, typically \( k \), which represents any integer.
To find the general solution of \( \cos(3x) = \frac{1}{2} \):

• The cosine function equals \( \frac{1}{2} \) at angles \( \theta = \frac{\pi}{3} \) and \( \theta = -\frac{\pi}{3} \) within one complete cycle.
• Accounting for full cycles, we express this as \( 3x = \frac{\pi}{3} + 2k\pi \) and \( 3x = -\frac{\pi}{3} + 2k\pi \).

These equations capture every instance where the cosine function achieves the value \( \frac{1}{2} \), enabling us to derive solutions by adjusting \( k \). This is why the general solution is crucial, as it provides a formulaic way to generate all possible solutions of the trigonometric equation.

Interval [0, 2π)

When solving trigonometric equations, the interval \([0, 2\pi)\) refers to the range of angles we are interested in considering. This interval includes all the angles from 0 up to, but not including, \(2\pi\).
It is crucial in trig problems because:

• It represents one full cycle of the trigonometric functions.
• Simplifies problems by restricting solutions to the most relevant range.
• Adds context to where solutions fall on the unit circle.

When we solve trigonometric equations within this interval, like in our exercise, we first derive the general solutions and then check which specific solutions actually fall within this range.
By testing different integer values for \( k \), we manage to identify the subset of solutions that meet our requirements of being within \([0, 2\pi)\). This narrows down the infinite solutions found in the full general solution to just the practical ones for our context.

Exact Solutions

Finding exact solutions in trigonometry involves determining the precise angles that satisfy the given equation within a specified interval, often \([0, 2\pi)\). These solutions are not approximations and are typically expressed in terms of fractions of \( \pi \).
The process for finding these involves:

• Calculating the general solution for the equation, as discussed previously.
• Substituting different values of \( k \) to find distinct solutions.
• Selecting the solutions that fall within the desired interval.

For example, from our equation \( \cos(3x) = \frac{1}{2} \), and with formulas derived, we found the valid solutions: \( \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}, \frac{11\pi}{9}, \frac{13\pi}{9}, \frac{17\pi}{9} \).
These are considered exact because they precisely meet the criteria set by the initial equation and specified interval, ensuring accuracy in their representation on the unit circle. This aspect is vital for problem-solving in mathematics, where precision is often required.