Problem 19 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

\(x = \frac{\pi}{3} + 2k\pi\) and \(x = \frac{5\pi}{3} + 2k\pi\), for any integer \(k\).

Step 1: Recognize the Quadratic Form

Notice that the given equation \(4 \cos^2 x - 4 \cos x + 1 = 0\) is in quadratic form where \( \cos x \) is the variable. This means we can treat it like a quadratic equation of the form \(4y^2 - 4y + 1 = 0\) with \(y = \cos x\).

Step 2: Solve the Quadratic Equation

Rewrite the quadratic equation in standard form: \(4y^2 - 4y + 1 = 0\). We can apply the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -4\), and \(c = 1\). Calculating the discriminant, \(b^2 - 4ac = (-4)^2 - 4 \cdot 4 \cdot 1 = 0\). Since the discriminant is zero, there is exactly one solution for \(y\).

Step 3: Apply the Quadratic Formula

Using the quadratic formula with the discriminant as zero, we find:\[ y = \frac{-(-4) \pm \sqrt{0}}{8} = \frac{4}{8} = \frac{1}{2} \]Therefore, \(\cos x = \frac{1}{2}\).

Step 4: Find All Solutions for \(\cos x = \frac{1}{2}\)

The equation \(\cos x = \frac{1}{2}\) is solved for \(x\) in the interval \([0, 2\pi]\). Solutions occur at:\[ x = \frac{\pi}{3}, \quad x = \frac{5\pi}{3} \]Remember that cosine is periodic with period \(2\pi\), so solutions for \(x\) in general are:\[ x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{3} + 2k\pi \]where \(k\) is any integer.

Step 5: State the Final Answer in General Form

The solutions for the equation \(4 \cos^2 x - 4 \cos x + 1 = 0\) are:\[ x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{3} + 2k\pi \] where \(k\) is any integer which indicates all possible solutions based on the periodic nature of cosine.

Key Concepts

Quadratic Form

In trigonometric equations, it's not uncommon to encounter expressions that can be likened to quadratic equations. A quadratic equation typically takes the form of \(ax^2 + bx + c = 0\), where \(x\) is the variable, and \(a, b,\) and \(c\) are constants.
In the exercise, the equation \(4 \cos^2 x - 4 \cos x + 1 = 0\) is structured like a quadratic equation. Here, \(\cos x\) acts as the variable. This allows us to treat it like \(4y^2 - 4y + 1 = 0\) by substituting \(y = \cos x\).
This technique of recognizing and reformulating the problem in a quadratic form simplifies the process and makes it easier to solve for the unknowns.

Cosine Function

The cosine function, denoted as \(\cos(x)\), is a fundamental trigonometric function. It tells us about the horizontal coordinate of a point on the unit circle as it travels counterclockwise from the starting position.
For example, \(\cos(0) = 1\), \(\cos\left(\frac{\pi}{2}\right) = 0\), and \(\cos(\pi) = -1\).

• The cosine function ranges from -1 to 1, meaning any value of \(\cos(x)\) will always be between these two extremes.
• It is periodic, repeating its values in a consistent pattern over a specific interval.

In the equation given in the exercise, solving \(\cos(x) = \frac{1}{2}\) leads us to specific angles where this condition is true within one full cycle of \([0, 2\pi]\).

Periodicity

Periodicity refers to the repeating nature of trigonometric functions, and the cosine function, in particular, has a natural period of \(2\pi\). This means that the values of \(\cos(x)\) will repeat every \(2\pi\) radians.
Therefore, if \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\), this will also be true for values like \(\frac{\pi}{3} + 2\pi, \frac{\pi}{3} + 4\pi\), etc. Similarly, it holds for \(\cos\left(\frac{5\pi}{3}\right)\).

• For trigonometric equations, this periodic nature is crucial for finding all possible solutions, as it shows that patterns repeat indefinitely.
• Whenever you find a solution, additional solutions can be found by adding integer multiples of the period to the original solution.

General Solution

The concept of a general solution in trigonometry involves finding all possible angles that satisfy a given equation using the periodicity of trigonometric functions.
The general solution for \(\cos(x) = \frac{1}{2}\) was found to be
\[x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{3} + 2k\pi\] where \(k\) is any integer. This represents all the points along the circumference of the unit circle that will have a cosine value of \(\frac{1}{2}\).

• This formulation allows you to account for the infinite number of solutions caused by the periodic nature.
• By varying \(k\), which can take any integer value, an endless number of solutions aligned with this pattern can be established.

By understanding the general solution, you ensure that no angle is left out in cyclic trigonometric phenomena.