Problem 1 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find all \(x\) between 0 and \(2 \pi\) such that (a) \(4 \cos ^{2} x=3\). (b) \(2 \sin ^{2} x-\sin x-1=0 . \quad\) (Hint: this is a quadratic in \(\sin x .\) ) (c) \(\sin x=2 / 3\). (Give exact answers, as well as numerical approximations.)

Step-by-Step Solution

Answer

The solutions for \(x\) in the interval [0, \(2 \pi\)] are: \n(a) \(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\) \n(b) \(x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}\) \n(c) \(x = \arcsin(\frac{2}{3}), \pi - \arcsin(\frac{2}{3})\) approximately \(x = 0.7297, 2.4119\) radians.

Step 1: Solve \(4 \cos^2 x = 3\)

First, we will solve the first equation for \(x\). To do this, we divide both sides by 4, resulting in \(\cos^2x = \frac{3}{4}\). Taking the square root of both sides (and remembering that this could result in a positive or negative solution), we get \(\cos x = \pm \sqrt{\frac{3}{4}}\) or \(\cos x = \pm \frac{\sqrt{3}}{2}\). Using the unit circle, we find that solutions between 0 and \(2 \pi\) for \(\cos x = \frac{\sqrt{3}}{2}\) are \(x = \frac{\pi}{6}, \frac{11\pi}{6}\), and for \(\cos x = -\frac{\sqrt{3}}{2}\) are \(x = \frac{5\pi}{6}, \frac{7\pi}{6}\).

Step 2: Solve \(2 \sin^2 x - \sin x - 1 = 0\)

This equation is in standard quadratic form, where \(a =2\), \(b = -1\), and \(c = -1\). By applying the quadratic formula, \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\), the roots are \(\sin x =1\) and \(\sin x = -0.5\). Looking at the unit circle, we find the solutions of \(x\) that satisfy this equation are \(x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}\).

Step 3: Solve \(\sin x = \frac{2}{3}\)

This is a relatively straightforward equation to solve. Since the equation asks where \(\sin(x)\) is exactly \(\frac{2}{3}\), all that needs to be done is to take the arcsine (or inverse sine) of \(\frac{2}{3}\) to find \(x\). Function \(\sin(x)\) is \(\frac{2}{3}\) at \(x = \arcsin(\frac{2}{3}), \pi - \arcsin(\frac{2}{3})\). Numerically these angles are approximately \(x = 0.7297, 2.4119\) radians. Note that both solutions are within the interval from 0 to \(2 \pi\).

Key Concepts

Understanding the Unit Circle

The unit circle is a key tool in trigonometry, particularly when solving trigonometric equations. It represents a circle with a radius of one, centered at the origin of the coordinate plane. Each point on the circle corresponds to an angle in radians, providing a geometric way to understand trigonometric functions. When we say

• \( \cos x = \frac{\sqrt{3}}{2} \)
• \( \cos x = -\frac{\sqrt{3}}{2} \)

these correspond to specific angles found using the unit circle.
For example, angles like \( \frac{\pi}{6} \) and \( \frac{11\pi}{6} \) arise from the positive cosine value, while \( \frac{5\pi}{6} \) and \( \frac{7\pi}{6} \) relate to the negative value. Understanding how these angles are derived from the unit circle is crucial for mastering this segment of trigonometry.

Applying the Quadratic Formula

The quadratic formula is a powerful tool for solving polynomial equations of the form \( ax^2 + bx + c = 0 \). In trigonometry, we can apply this to functions of sine or cosine by considering them as variables. Here, when solving \[ 2 \sin^2 x - \sin x - 1 = 0 \]treat \( \sin x \) as a variable \( y \), transforming the equation into a standard quadratic form. By using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]we determine the possible values for \( \sin x \), which lead us to the angles specified by the unit circle. This methodological approach allows us to transition smoothly from algebra to trigonometry, finding exact solutions for the angles.

Exploring Inverse Sine

Inverse trigonometric functions help us uncover angles when given a specific trigonometric value. For sine, the inverse function is arcsin, denoted as \( \arcsin(x) \). It answers the question, "What angle has this sine value?" In \[ \sin x = \frac{2}{3} \]we find \( x = \arcsin\left(\frac{2}{3}\right) \) and \( x = \pi - \arcsin\left(\frac{2}{3}\right) \).
This function is particularly useful when the sine function doesn't match a standard angle from the unit circle. Exact solutions alongside numerical approximations provide essential insight, confirming that our answer falls within the specified interval.

Working Within the Interval [0, 2π]

When solving trigonometric equations, it's crucial to consider the interval within which solutions are valid. Typically, we operate within

• \([0, 2\pi] \)

representing one full rotation of the unit circle. Every solution we find should lie within this range, ensuring relevance to the given problem.
By understanding this interval, we align our solutions with standard conventions in trigonometry. It helps students recognize that trigonometric functions are periodic, repeating every \( 2\pi \) radians. This understanding is particularly vital for ensuring that solutions to trigonometric problems are applied correctly.