Problem 46 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 39-46: Refer to the graph of \(y=\sin x\) or \(y=\cos x\) to find the exact values of \(x\) in the interval \([0,4 \pi]\) that satisfy the equation. $$ \cos x=-\frac{1}{2} $$

Step-by-Step Solution

Answer

The exact values of \( x \) are \( \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3} \).

Step 1: Understand the Equation

The given equation is \( \cos x = -\frac{1}{2} \). We need to find the values of \( x \) in the interval \([0, 4\pi]\) that satisfy this condition.

Step 2: Identify the General Solutions

Consider the unit circle where \( \cos \theta = -\frac{1}{2} \). This occurs at an angle of \( \theta = \frac{2\pi}{3} \) and \( \theta = \frac{4\pi}{3} \) since these are the angles on the unit circle within \([0,2\pi]\) where cosine equals \(-\frac{1}{2}\).

Step 3: Extend the Solutions to the Interval

Since trigonometric functions are periodic, extend the solutions to the interval \([0, 4\pi]\), which includes two cycles of \(2\pi\). The solutions are \( x = \frac{2\pi}{3} + 2\pi k \) and \( x = \frac{4\pi}{3} + 2\pi k \), where \( k \) is an integer.

Step 4: Calculate Specific Values in the Interval

Calculate the specific values for \( k = 0, 1, 2 \) to find which values lie within \([0, 4\pi]\):- For \( k = 0 \): \( x = \frac{2\pi}{3}, \frac{4\pi}{3} \)- For \( k = 1 \): \( x = \frac{2\pi}{3} + 2\pi = \frac{8\pi}{3}, \frac{4\pi}{3} + 2\pi = \frac{10\pi}{3} \)- For \( k = 2 \): \( x = \frac{2\pi}{3} + 4\pi \) exceeds \( 4\pi \), similarly \( x = \frac{4\pi}{3} + 4\pi \) exceeds \( 4\pi \).

Step 5: Compile the Exact Values

Considering above calculations, the exact values of \( x \) within \([0, 4\pi]\) that satisfy \( \cos x = -\frac{1}{2} \) are \( \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3} \).

Key Concepts

Unit Circle

The unit circle is a fundamental concept in trigonometry, used to define the sine and cosine of an angle. It is a circle with a radius of 1, centered at the origin (0, 0) of a coordinate plane. This circle is particularly useful when analyzing trigonometric functions because the x-coordinate of any point on the circle corresponds to the cosine of an angle, while the y-coordinate corresponds to the sine. This makes it simple to visualize and determine values for angles.

• Each point on the unit circle represents a complex number associated with an angle measured from the positive x-axis.
• Angles can be measured in radians, where one full revolution around the circle is equal to \(2\pi\) radians.
• The unit circle helps find exact values of trigonometric functions without needing a calculator.

For our exercise, if we place points that align with \(\cos \theta = -\frac{1}{2}\), we look for angles where the x-coordinate equals \(-\frac{1}{2}\). We find these values at \(\theta = \frac{2\pi}{3}\) and \(\theta = \frac{4\pi}{3}\) on the unit circle.

Cosine Function

The cosine function is an important trigonometric function that shows the relation between the angle and length of the adjacent side in a right triangle. In the context of the unit circle, the cosine of an angle is the x-coordinate of the point where the terminal side of the angle intersects the circle.

• The cosine function is denoted as \(\cos(x)\) and maps real numbers (angles in radians) to real values ranging between -1 and 1.
• The graph of the cosine function is a wave that starts at 1 when \(x=0\) and periodically repeats every \(2\pi\).
• The function decreases to 0 at \(\pi/2\), to -1 at \(\pi\), back to 0 at \(3\pi/2\), and returns to 1 at \(2\pi\).

The exercise requires identifying points where \(\cos(x) = -\frac{1}{2}\). Referring to the unit circle and graph of the cosine function, these points are \(x = \frac{2\pi}{3}\) and \(x = \frac{4\pi}{3}\) within \([0, 2\pi]\).

Periodic Functions

Periodic functions are functions that repeat their values at regular intervals. In trigonometry, sine and cosine functions are classic examples of periodic functions with a fundamental period of \(2\pi\), meaning they repeat themselves every \(2\pi\) radians.

• The main property of periodic functions is their ability to continue infinitely with repetition—this makes them incredibly useful in modeling cyclical phenomena.
• The cosine function repeats every \(2\pi\), which allows us to extend solutions beyond the initial cycle.
• For the equation \(\cos x = -\frac{1}{2}\), we find the initial solutions within one cycle \([0, 2\pi]\) and duplicate these with an offset of \(2\pi\).

In our problem, given the interval \([0, 4\pi]\), we apply periodicity to find solutions that lie not only in the first cycle but also in the extended interval, resulting in additional solutions at \(x = \frac{8\pi}{3}\) and \(x = \frac{10\pi}{3}\).