Problem 48 - Trigonometry

Question

\(10 \cos ^{2} \theta+\cos \theta-3=0\)

Step-by-Step Solution

Answer

\( \theta = \frac{\pi}{3}, \frac{5\pi}{3} \) or \( \theta \approx \cos^{-1}(-3/5), 2\pi-\cos^{-1}(-3/5) \).

Step 1: Identify the Type of Equation

The given equation is in the form of a quadratic in terms of \( \cos \theta \). It is \( 10 \cos^2 \theta + \cos \theta - 3 = 0 \). Our goal is to solve for \( \theta \).

Step 2: Substitute for Simplicity

Let \( x = \cos \theta \). Substitute this into the equation to get \( 10x^2 + x - 3 = 0 \). Now we have a quadratic equation in terms of \( x \).

Step 3: Apply the Quadratic Formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For our equation, \( a = 10 \), \( b = 1 \), \( c = -3 \). Substitute these values into the formula.

Step 4: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac = 1^2 - 4 \cdot 10 \cdot (-3) = 1 + 120 = 121 \). The discriminant is positive, indicating two distinct real solutions for \( x \).

Step 5: Find the Roots of the Quadratic

Plug the discriminant into the quadratic formula: \[x = \frac{-1 \pm \sqrt{121}}{20}\]This gives two solutions: \( x = \frac{-1 + 11}{20} = \frac{10}{20} = \frac{1}{2} \) and \( x = \frac{-1 - 11}{20} = \frac{-12}{20} = -\frac{3}{5} \).

Step 6: Translate Back to Trigonometric Form

Now substitute back, noting that \( x = \cos \theta \).1. \( \cos \theta = \frac{1}{2} \)2. \( \cos \theta = -\frac{3}{5} \).

Step 7: Solve for \( \theta \)

Solve each equation:1. For \( \cos \theta = \frac{1}{2} \), \( \theta = \frac{\pi}{3} + 2k\pi \) or \( \theta = \frac{5\pi}{3} + 2k\pi \) for any integer \( k \).2. For \( \cos \theta = -\frac{3}{5} \), use a calculator or table:\( \theta \approx \cos^{-1}\left( -\frac{3}{5} \right) \) gives the principal angle, with the second solution being \( 2\pi - \cos^{-1}\left( -\frac{3}{5} \right) \).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and are solvable using several methods, including factoring, completing the square, and the quadratic formula. A quadratic equation is usually written in the standard form:

• \( ax^2 + bx + c = 0 \)

Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). The simple structure of quadratic equations makes them versatile for solving a variety of problems.
Because they are polynomials of degree 2, quadratic equations generally have two solutions, which can be real or complex numbers. Understanding this concept is key, as seen in trigonometric equations transformed into quadratic ones, like in our exercise.

Cosine Function

The cosine function, denoted as \( \cos(x) \), is one of the main functions in trigonometry and is crucial in solving trigonometric equations. It represents the x-coordinate of the point on a unit circle corresponding to a given angle, measured in radians.
For any angle \( \theta \), the cosine function can take values between -1 and 1, which gives it distinctive properties essential in many fields, including physics and engineering.

• The cosine function is periodic, with a period of \( 2\pi \).
• It is an even function, meaning \( \cos(-\theta) = \cos(\theta) \).

In trigonometric equations like the one given, transforming the expression to a quadratic form simplifies finding the cosine values of \( \theta \), making the equation easier to solve.

Quadratic Formula

The quadratic formula is a powerful tool for finding the roots of any quadratic equation. It is expressed as:

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

This formula provides a systematic way to find the solutions of quadratic equations, even when other methods like factoring are not applicable.
The quadratic formula relies on calculating the discriminant, \( b^2 - 4ac \), which reveals the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If zero, there is exactly one real root (a repeated root).
• If negative, there are two complex roots.

Incorporating the quadratic formula for solving an equation involving the cosine function, as in our problem, allows us to find the different cosine values, moving from an algebraic to a trigonometric solution.