Problem 18 - Algebra 2

Question

Solve each equation for all values of ? if ? is measured in radians. \(\cos ^{2} \theta-\frac{5}{2} \cos \theta-\frac{3}{2}=0\)

Step-by-Step Solution

Answer

\( \theta = \frac{2\pi}{3} \) and \( \theta = \frac{4\pi}{3} \)

Step 1: Identify the Type of Equation

Recognize that the equation \( \cos^2 \theta - \frac{5}{2} \cos \theta - \frac{3}{2} = 0 \) is a quadratic equation in terms of \( \cos \theta \). Let's set \( x = \cos \theta \) for simplification, giving us the quadratic equation \( x^2 - \frac{5}{2}x - \frac{3}{2} = 0 \).

Step 2: Apply the Quadratic Formula

To solve the quadratic equation, we use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = 1 \), \( b = -\frac{5}{2} \), and \( c = -\frac{3}{2} \). Substitute these values into the formula:

Step 3: Calculate the Discriminant

Calculate the discriminant \( \Delta = b^2 - 4ac \):\[ \Delta = \left(-\frac{5}{2}\right)^2 - 4 \cdot 1 \cdot \left(-\frac{3}{2}\right) = \frac{25}{4} + 6 = \frac{49}{4} \]

Step 4: Solve for \( x \)

Substitute the discriminant back into the quadratic formula:\[ x = \frac{-(-\frac{5}{2}) \pm \sqrt{\frac{49}{4}}}{2} = \frac{\frac{5}{2} \pm \frac{7}{2}}{2} \]This gives us two solutions:1. \( x = \frac{5/2 + 7/2}{2} = 3 \)2. \( x = \frac{5/2 - 7/2}{2} = -\frac{1}{2} \)

Step 5: Determine Solutions in Terms of \( \theta \)

Since \( x = \cos \theta \), recall that the range of \( \cos \theta \) is [-1, 1]. Thus, only \( x = -\frac{1}{2} \) is valid;- For \( \cos \theta = -\frac{1}{2} \), \( \theta = \frac{2\pi}{3} \) and \( \theta = \frac{4\pi}{3} \) based on the unit circle.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two that are usually in the form of \( ax^2 + bx + c = 0 \). These types of equations are fundamental in algebra and have various methods of solving them. For instance, one can factor the equation, complete the square, or use the quadratic formula. The quadratic formula is commonly used when an equation cannot be easily factored. It is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In the context of trigonometric equations, terms like \( \cos^2 \theta \) serve as the variable \( x \), allowing the equation to be treated as a quadratic. This approach simplifies the complexity by converting a trigonometric equation into a more familiar quadratic equation.Recognizing and setting up the equation appropriately is key, just as in the provided example where we substitute \( x = \cos \theta \) to transform the trigonometric equation into a quadratic form. This simplification makes the solving process more straightforward.

Trigonometric Functions

Trigonometric functions are mathematical functions related to angles. In trigonometry, the most common functions are sine, cosine, and tangent, often represented as \( \sin \theta \), \( \cos \theta \), and \( \tan \theta \). Each function corresponds to specific ratios derived from a right-angled triangle. Cosine, for example, is the ratio of the adjacent side to the hypotenuse when working with angles in a right-angled triangle.
These functions are periodic, meaning they repeat their values in regular intervals, and are often used to model cyclical behaviors. In trigonometric equations, these functions are solved for specific angle values, usually within a defined range, using relationships such as the Pythagorean identity. Such techniques allow us to find the angle solutions that the trigonometric function may correspond to within a given interval. For instance, knowing that \( \cos \theta = -\frac{1}{2} \) and utilizing the unit circle can help us find exact angles that satisfy this condition.Trigonometric functions also rely heavily on certain identities and properties, such as their ranges and periodicity. This means acknowledging that \( \cos \theta \) has a range from -1 to 1, influencing the solutions for any equation it's a part of, as demonstrated in converting the quadratic roots into meaningful trigonometric solutions.

Unit Circle

The unit circle is a powerful tool when dealing with trigonometric functions. It is a circle with a radius of 1 centered at the origin of the coordinate system. This geometric representation allows us to easily determine the values of trigonometric functions for any given angle. Each point on the unit circle corresponds to coordinates \( (\cos \theta, \sin \theta) \). These coordinates directly give the values of the cosine and sine of that angle.
Using the unit circle, one can find the angle solutions to trigonometric equations. For example, when solving \( \cos \theta = -\frac{1}{2} \), the unit circle reveals that the relevant angles are \( \frac{2\pi}{3} \) and \( \frac{4\pi}{3} \). These angles are the points on the circle where the x-coordinate, representing \( \cos \theta \), equals -\frac{1}{2}. Utilizing this visual and conceptual tool significantly aids in the understanding of trigonometric solutions and their periodic nature.The unit circle also helps in visualizing the periodicity of trigonometric functions, which is crucial for understanding how functions like sine and cosine repeat after full rotations of \( 2\pi \) radians. This repetition enables the determination of additional solutions by adding full circle revolutions to the initial answer.