Problem 41 - Algebra and Trigonometry

Question

Find all solutions of the equation in the interval \([0,2 \pi).\) $$2 \cos 3 x=1$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{\pi}{9}, \frac{7\pi}{9}, \frac{13\pi}{9}, \frac{19\pi}{9}, \frac{5\pi}{9}, \frac{11\pi}{9}, \frac{17\pi}{9}\) in the interval \([0, 2\pi)\).

Step 1: Isolate the Trigonometric Function

We start with the equation \(2 \cos 3x = 1\). To make it easier to work with, divide both sides of the equation by 2 to get \(\cos 3x = \frac{1}{2}\).

Step 2: Find the General Solutions

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. Thus, for \(3x\), we have two sets of solutions: \(3x = \frac{\pi}{3} + 2k\pi\) and \(3x = -\frac{\pi}{3} + 2k\pi\).

Step 3: Solve for x

Divide both sides of the equations from the previous step by 3 to solve for \(x\). This gives two sets of general solutions for \(x\): \(x = \frac{\pi}{9} + \frac{2k\pi}{3}\) and \(x = -\frac{\pi}{9} + \frac{2k\pi}{3}\).

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

Determine the specific solutions of \(x\) from the general forms by plugging in integer values for \(k\) such that \(0 \leq x < 2\pi\).- For the equation \(x = \frac{\pi}{9} + \frac{2k\pi}{3}\): - \(k = 0\) gives \(x = \frac{\pi}{9}\). - \(k = 1\) gives \(x = \frac{7\pi}{9}\). - \(k = 2\) gives \(x = \frac{13\pi}{9}\). - \(k = 3\) gives \(x = \frac{19\pi}{9}\) which is within the interval. - \(k = 4\) gives \(x = \frac{25\pi}{9}\) which is outside the interval.- For the equation \(x = -\frac{\pi}{9} + \frac{2k\pi}{3}\): - \(k = 0\) gives \(x = -\frac{\pi}{9}\) which is outside the interval. - \(k = 1\) gives \(x = \frac{5\pi}{9}\). - \(k = 2\) gives \(x = \frac{11\pi}{9}\). - \(k = 3\) gives \(x = \frac{17\pi}{9}\). - \(k = 4\) gives \(x = \frac{23\pi}{9}\) which is outside the interval.

Key Concepts

Trigonometric Function

Trigonometric functions are essential in mathematics and are often used to describe periodic phenomena. These functions, including sine and cosine, depend on angles and are characterized by their repeating values. The cosine function, denoted as \( \cos \theta \), outputs values ranging from -1 to 1. This wide range makes the cosine function very useful in solving equations related to rotation, waves, and oscillations.
When we deal with equations involving trigonometric functions, our goal is often to find the angles that yield a particular value of the function. For example, in the problem given, we have the equation \( 2 \cos 3x = 1 \). By simplifying, we rearrange it into a basic cosine equation \( \cos 3x = \frac{1}{2} \). Such equations often appear in algebraic and geometric problems related to angles and rotations.

General Solutions

The general solution of a trigonometric equation is a comprehensive representation of all possible solutions. These solutions consider the periodic nature of trigonometric functions, recognizing that they repeat their values at regular intervals.
In our problem, we first isolated the cosine function, then identified the initial angles from the unit circle for which \( \cos \theta = \frac{1}{2} \). These angles are \( \frac{\pi}{3} \) and \( -\frac{\pi}{3} \). However, due to the periodic nature of the cosine function, these solutions repeat every \( 2\pi \). Therefore, in terms of general solutions for \( 3x \), we express them as:

• \( 3x = \frac{\pi}{3} + 2k\pi \)
• \( 3x = -\frac{\pi}{3} + 2k\pi \)

Here, \( k \) represents any integer, allowing us to spin around the circle an arbitrary number of times, thus capturing this periodicity.

Interval Solutions

Interval solutions refer to the specific solutions that satisfy the condition constrained within a given range, or interval. In the problem at hand, we are asked to find the solutions of \( x \) within the interval \([0, 2\pi)\).
After computing our general solutions for \( x \), which resulted in:

• \( x = \frac{\pi}{9} + \frac{2k\pi}{3} \)
• \( x = -\frac{\pi}{9} + \frac{2k\pi}{3} \)

We substitute integer values into \( k \) to calculate specific values of \( x \) within the desired interval, ensuring \( 0 \leq x < 2\pi \). This method involves testing successive integer values of \( k \) and checking which results meet the interval condition. This step is indispensable to ensure we consider only the solutions relevant to our specified interval.

Cosine Function

The cosine function is one of the fundamental trigonometric functions, typically defined using the unit circle. Any angle's cosine value corresponds to the x-coordinate of the point where the terminal side of the angle intersects the unit circle.
For the equation \( \cos 3x = \frac{1}{2} \), recognizing where \( \cos \theta = \frac{1}{2} \) is crucial. From trigonometric knowledge, cosine equals \( \frac{1}{2} \) at specific angles, namely \( \frac{\pi}{3} \) and \( 5\pi/3 \). However, because the cosine function repeats in cycles of \( 2\pi \), we must consider these initial angles within this cycle, reflecting in our general solution.
Being periodic, the cosine function enables finding multiple solutions for trigonometric equations, and understanding these inherent properties aids in solving such problems efficiently.