91Ó°ÊÓ

A quadratic equation is any equation that can be rearranged in standard form as \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). These equations are unique because they describe a parabolic curve on a graph.

In the given problem, we were able to rearrange the initial equation \( 8 \cos^2 \theta = 3 - 2 \cos \theta \) into the quadratic form \( 8 \cos^2 \theta + 2 \cos \theta - 3 = 0 \). By letting \( x = \cos \theta \), we obtain \( 8x^2 + 2x - 3 = 0 \), which is a standard quadratic equation.

Quadratic equations can usually be solved by:

• Factoring into two binomials
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Completing the square

For this exercise, factoring was used to find the solutions. This method finds factors that multiply to \( a \times c \) and add to \( b \). In our case, factoring \( 8x^2 + 2x - 3 \) yields \((4x - 1)(2x + 3) = 0 \). Each factor leads to possible solutions for \( x \) but one may need to be discarded if it yields an invalid trigonometric value.

The cosine function is one of the primary trigonometric functions and is denoted as \( \cos(\theta) \). It is the x-coordinate of the point where the terminal side of an angle \( \theta \) intersects the unit circle.

The function follows these properties:

• Periodicity: It has a period of \( 2\pi \), meaning it repeats its values every \( 2\pi \).
• Range: The values of \( \cos \theta \) are always between \(-1\) and \(1\).
• Even Function: It is symmetric about the y-axis, which means \( \cos(-\theta) = \cos(\theta) \).

In our exercise, we found that \( \cos \theta = \frac{1}{4} \), a valid value within the defined range. It is important to discard any \( x \) values outside the valid cosine range, like \( x = -\frac{3}{2} \), because they cannot represent the cosine of any real angle.

Inverse trigonometric functions allow us to find the angle that corresponds to a given trigonometric value. For cosine, we use \( \cos^{-1}(x) \), also known as arccosine.

The function \( \cos^{-1}(x) \) yields an angle \( \theta \) such that \( 0 \leq \theta \leq \pi \). This is called the principal value. However, since we are dealing with the interval \([0, 2\pi)\), we must consider angles in both first and fourth quadrants for cosine.

For our problem:

• The principal value is \( \theta = \cos^{-1}(\frac{1}{4}) \), which lies in the first quadrant.
• We also need the equivalent angle in the fourth quadrant: \( \theta = 2\pi - \cos^{-1}(\frac{1}{4}) \).

This way, we utilize inverse trigonometric functions to provide comprehensive solutions over a specified interval. Remember, exact solutions with inverse trig functions may not always result in simple, elementary numbers.