Problem 28 - Calculus with Concepts in Calculus

Question

Solve the equation \(4 \cos ^{2} x-4 \sqrt{3} \cos x+3=0\).

Step-by-Step Solution

Answer

\(x = \frac{\pi}{6} + 2k\pi\) and \(x = \frac{11\pi}{6} + 2k\pi\), where \(k \in \mathbb{Z}\).

Step 1: Identify the Type of Equation

The given equation \(4 \cos^{2} x - 4 \sqrt{3} \cos x + 3 = 0\) is a quadratic equation in terms of \(\cos x\). We can write it as \(a \cos^{2} x + b \cos x + c = 0 \) with \(a = 4\), \(b = -4\sqrt{3}\), and \(c = 3\).

Step 2: Apply the Quadratic Formula

To solve for \(\cos x\), use the quadratic formula, \(\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute \(a = 4\), \(b = -4\sqrt{3}\), and \(c = 3\) into the formula.

Step 3: Calculate the Discriminant

The discriminant is calculated as \(b^2 - 4ac\). Here, \(b = -4\sqrt{3}\) and \(c = 3\), so the discriminant is:\((-4\sqrt{3})^2 - 4 \times 4 \times 3 = 48 - 48 = 0\).

Step 4: Solve for \(\cos x\) Using Positive Discriminant

Since the discriminant is zero, there is only one solution for \(\cos x\). Substitute into the quadratic formula:\[\cos x = \frac{-(-4\sqrt{3})}{2 \times 4} = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}.\]

Step 5: Find the Angles for the Cosine Value

Solve \(\cos x = \frac{\sqrt{3}}{2}\), which corresponds to angles in standard position. In radians, \(x = \frac{\pi}{6}\) and \(x = \frac{11\pi}{6}\). In degrees, this translates to \(x = 30^\circ\) and \(x = 330^\circ\).

Step 6: General Solution

Since \(\cos\) function is periodic with a period of \(2\pi\), the general solutions will be:\[x = \frac{\pi}{6} + 2k\pi \text{ and } x = \frac{11\pi}{6} + 2k\pi, \text{ where } k \in \mathbb{Z}.\]

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of the second degree. They typically have the form:

• \(ax^2 + bx + c = 0\)

where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable.
In our problem, we face a quadratic equation with \(\cos x\) treating as the variable:

• \(4 \cos^2 x - 4 \sqrt{3} \cos x + 3 = 0\)

The constants \(a = 4\), \(b = -4\sqrt{3}\), and \(c = 3\) represent the coefficients in this equation.
Quadratic equations are solved by applying various methods, such as factoring, completing the square, or using the quadratic formula.
In our original exercise, we focus on using the quadratic formula to find the roots relevant to the cosine function.

Cosine Function

The cosine function, denoted as \(\cos x\), is one of the fundamental trigonometric functions. It describes the relationship between an angle in a right triangle and the ratio of the adjacent side to the hypotenuse.
The cosine function is also periodic with a period of \(2\pi\), meaning that:

• \(\cos(x) = \cos(x + 2k\pi)\) for any integer \(k\).

In our problem, we find that \(\cos x = \frac{\sqrt{3}}{2}\).
The cosine function reaches the value \(\frac{\sqrt{3}}{2}\) at specific standard angles:

• \(x = \frac{\pi}{6}\)
• \(x = \frac{11\pi}{6}\)

These solutions are found within the first cycle of one periodic interval of the cosine curve.

Trigonometric Solutions

Trigonometric solutions involve finding all angles that satisfy a given trigonometric equation.
The results we found, \(x = \frac{\pi}{6}\) and \(x = \frac{11\pi}{6}\), are particular solutions within a single period of the cosine function.
To find all possible solutions,

• We leverage the periodic nature of the cosine function using the formula:
  • \(x = \frac{\pi}{6} + 2k\pi\)
  • \(x = \frac{11\pi}{6} + 2k\pi\)
  where \(k\) is any integer.

These represent the general solutions, capturing the infinite number of angles satisfying the original equation, all spaced by the function's period of \(2\pi\).

Discriminant Calculation

In the quadratic equation, the discriminant helps to determine the nature of the roots.
The discriminant is calculated from the formula:

• \(b^2 - 4ac\)

In our exercise:

• The discriminant is \((-4\sqrt{3})^2 - 4 \times 4 \times 3\).
• This equals \(48 - 48 = 0\).

When the discriminant is zero, it indicates a single repeated solution for the quadratic equation.
Thus, in terms of our trigonometric equation, \(\cos x\) yields precisely one unique value, simplifying the process of finding associated angles.