Problem 47 - Precalculus Essentials

Question

Involve trigonometric equations quadratic in form. Solve each equation on the interval \([0,2 \pi)\) $$4 \cos ^{2} x-1=0$$

Step-by-Step Solution

Answer

The solution to the trigonometric equation is \( x = \frac{\pi}{3}, \frac{5 \pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3} \)

Step 1: Rearrange the equation

First, rearrange the given equation to form a standard quadratic equation. This can be done by moving the constant term to the right hand side of the equation. Hence, we get \(4 \cos ^{2} x = 1 \).

Step 2: Simplify the equation

The next step is to get rid of the coefficient of \(\cos ^{2} x\), which is currently 4, to simplify the equation. We can do this by dividing both sides of the equation by 4. This results in the simplified equation \( \cos ^{2} x =\frac{1}{4}\).

Step 3: Take the square root

Next, take the square root of both sides to get the cosine of \( x \). However, important to note that when square rooting, there will be two solutions: positive and negative. Hence, we get \( \cos x = \frac{1}{2}\) and \( \cos x = - \frac{1}{2} \).

Step 4: Solve for \( x \)

Now, find all \( x \) that satisfy \( \cos x = \frac{1}{2}\) and \( \cos x = - \frac{1}{2}\) in the specified interval ([0, \( 2 \pi \))). Upon solving, we know from the unit circle that \( x = \frac{\pi}{3}, \frac{5 \pi}{3}\) for \( \cos x = \frac{1}{2}\) and \( x = \frac{2 \pi}{3}, \frac{4 \pi}{3}\) for \( \cos x = - \frac{1}{2}\).

Key Concepts

Quadratic Equations

In trigonometric equations, similar to algebraic ones, quadratic equations can appear. A quadratic equation is usually in the form \(ax^2 + bx + c = 0\). In the given exercise, the trigonometric equivalent is \(4 \cos^2 x - 1 = 0\). Here, \(\cos^2 x\) acts like \(x^2\) does in algebraic equations.
To solve these equations, it is first necessary to convert them to the standard quadratic form and then apply solution techniques that you would use for regular quadratic equations.

• Simplification involves rearranging terms to an equivalent version with a known structure.
• The equation \(4 \cos^2 x = 1\) is analogous to isolating \(ax^2\) by moving constants across the equals sign.
• Next, we divide through by the leading coefficient, similar to obtaining a simplified \(x^2\)-type expression in algebra.

Understanding these manipulations is key to recognizing and solving trigonometric expressions that follow a quadratic form.

Unit Circle

The unit circle is a fundamental concept in trigonometry. It is a circle with a radius of 1 centered at the origin on the coordinate plane. It is essential for understanding how angles and trigonometric functions relate.
For the exercise, you need to find where on the unit circle the value of \(\cos(x) = \frac{1}{2}\) and \(\cos(x) = -\frac{1}{2}\). These correspond to specific angles:

• \(\cos(x) = \frac{1}{2}\) occurs at \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\).
• \(\cos(x) = -\frac{1}{2}\) occurs at \(x = \frac{2\pi}{3}\) and \(x = \frac{4\pi}{3}\).

These positions are derived by identifying where on the unit circle these cosine values are met. In each case, you're dealing with symmetrical positions relative to the central axis (y-axis), which helps deduce these relationships.
This symmetry is invaluable for analytical problem-solving in trigonometry.

Interval Notation

Interval notation is a way of expressing the set of solutions for an equation within a specific range of values, which is crucial in solving trigonometric equations, as they often have infinite solutions. In this problem, we focus on the interval \([0, 2\pi)\).
This interval dictates that we consider all solutions "wrapping around" the unit circle once, starting from 0 and excluding \(2\pi\) itself, since \([0, 2\pi)\) represents all angles from 0 up to, but not including, 2 radians measured counterclockwise.

• Such constraints ensure we consider only unique solutions within a defined range.
• Interval notation typically uses brackets \([ ]\) for inclusive and parentheses \(( )\) for inclusive, making it a powerful tool for precise definition.

Applying these boundaries helps in deriving responses suitable for one complete rotation, matching the context of periodic functions evaluated over the unit circle.