Problem 7 - Algebra and Trigonometry

Question

Find all solutions of the equation. $$4 \cos ^{2} x-1=0$$

Step-by-Step Solution

Answer

All solutions are \( x = \frac{\pi}{3} + 2k\pi, \; x = -\frac{\pi}{3} + 2k\pi, \; x = \frac{2\pi}{3} + 2k\pi, \; x = -\frac{2\pi}{3} + 2k\pi, \; k \in \mathbb{Z} \).

Step 1: Identify the Equation Type

The given equation is a trigonometric equation involving the cosine function. It is specifically a quadratic equation in terms of \( \cos x \).

Step 2: Rearrange the Equation

We start by rearranging the equation:\[ 4 \cos^2 x - 1 = 0 \]Add 1 to both sides:\[ 4 \cos^2 x = 1 \]

Step 3: Solve for \( \cos^2 x \)

Divide both sides by 4 to isolate \( \cos^2 x \):\[ \cos^2 x = \frac{1}{4} \]

Step 4: Solve for \( \cos x \)

Take the square root of both sides to solve for \( \cos x \):\[ \cos x = \pm \frac{1}{2} \]

Step 5: Find the General Solutions for \( \cos x = \frac{1}{2} \)

The solutions to \( \cos x = \frac{1}{2} \) are given by the angles where the cosine has a value of \( \frac{1}{2} \):\[ x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = -\frac{\pi}{3} + 2k\pi, \; k \in \mathbb{Z} \]

Step 6: Find the General Solutions for \( \cos x = -\frac{1}{2} \)

The solutions to \( \cos x = -\frac{1}{2} \) are given by the angles where the cosine has a value of \(-\frac{1}{2} \):\[ x = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad x = -\frac{2\pi}{3} + 2k\pi, \; k \in \mathbb{Z} \]

Step 7: Combine All Solutions

Combine the solutions from steps 5 and 6:\[ x = \frac{\pi}{3} + 2k\pi, \; x = -\frac{\pi}{3} + 2k\pi, \; x = \frac{2\pi}{3} + 2k\pi, \; x = -\frac{2\pi}{3} + 2k\pi, \; \text{where} \; k \in \mathbb{Z} \]

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and appear commonly in various mathematical scenarios. A standard quadratic equation has the form: \[ ax^2 + bx + c = 0 \]where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). In the exercise provided, the equation \(4 \cos^2 x - 1 = 0\) is a quadratic equation rearranged in terms of \(\cos x\).

• The term \(4 \cos^2 x\) mimics \(ax^2\).
• The constant \(-1\) mimics \(c\) in the standard formula.

To solve such equations, we often rearrange them to find the roots, i.e., the values of \(x\) that satisfy the equation. This might involve factoring, using the quadratic formula, or, as in this case, transforming it into a simpler form by using algebraic manipulations.

Cosine Function

The cosine function is one of the primary trigonometric functions. It is commonly associated with right-angled triangles and the unit circle. For any angle \(x\), the cosine function \(\cos x\) represents the adjacent side of a right triangle over its hypotenuse.
Additionally, on the unit circle, \(\cos x\) corresponds to the x-coordinate of the point where the terminal side of the angle intersects the circle. This cyclical behavior makes trigonometric equations quite interesting.
In the original problem, the involvement of the cosine function transforms a simple quadratic equation into a trigonometric one. Solving \(\cos x = \pm \frac{1}{2}\) requires considering all potential angles \(x\) where the cosine function delivers those values.

• Cosine equals \(\frac{1}{2}\) at typical angles like \(\frac{\pi}{3}\) and others.
• Cosine equals \(-\frac{1}{2}\) at angles like \(\frac{2\pi}{3}\).

General Solutions

The concept of general solutions is crucial in trigonometry. Trigonometric functions, like cosine, are periodic, meaning they repeat values in regular intervals. This periodic nature requires us to express solutions in a general form, accounting for all possible angles providing the same cosine value.
To define these general solutions, we express equations with an additional term representing the periodicity, often using multiples of \(2\pi\), as the cosine function cycles every \(2\pi\) radians. For instance:

• \(x = \frac{\pi}{3} + 2k\pi\) reflects one such general solution.
• The term \(2k\pi\) accounts for the infinite repetitions of \(\cos x = \frac{1}{2}\) or \(-\frac{1}{2}\).

Here, \(k\) is an integer, accommodating every complete cycle (or rotation) been made to return to an equivalent angle, ensuring no potential angle is overlooked. General solutions provide a comprehensive understanding of where and how often a specific trigonometric condition is met across infinite cycles.