Problem 17 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation in part (a) analytically over the interval \([0,2 \pi) .\) Then use a graph to solve each inequality in part (b). (a) \(\sqrt{2} \cos 2 x=-1\) (b) \(\sqrt{2} \cos 2 x \leq-1\)

Step-by-Step Solution

Answer

The solutions to the equation are \( x = \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8} \). The inequality holds at these points.

Step 1: Solve the Equation

To solve the equation \( \sqrt{2} \cos 2x = -1 \), first divide both sides by \( \sqrt{2} \) to isolate the cosine term:\[ \cos 2x = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \]Now, find the general solutions for \( 2x \) such that \( \cos 2x = -\frac{\sqrt{2}}{2} \). The angles where this occurs are \( 2x = \frac{3\pi}{4} + 2k\pi \) and \( 2x = \frac{5\pi}{4} + 2k\pi \), where \( k \) is an integer.

Step 2: Solve for x

From the equations \( 2x = \frac{3\pi}{4} + 2k\pi \) and \( 2x = \frac{5\pi}{4} + 2k\pi \), solve for \( x \):1. \( x = \frac{3\pi}{8} + k\pi \)2. \( x = \frac{5\pi}{8} + k\pi \)Determine the values of \( k \) such that \( x \) stays within \( [0, 2\pi) \):- For \( x = \frac{3\pi}{8} + k\pi \): - \( k = 0 \) gives \( x = \frac{3\pi}{8} \) - \( k = 1 \) gives \( x = \frac{11\pi}{8} \)- For \( x = \frac{5\pi}{8} + k\pi \): - \( k = 0 \) gives \( x = \frac{5\pi}{8} \) - \( k = 1 \) gives \( x = \frac{13\pi}{8} \)

Step 3: Summary of Analytical Solutions

The solutions to the equation \( \sqrt{2} \cos 2x = -1 \) in the interval \( [0, 2\pi) \) are \( x = \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8} \).

Step 4: Graph and Solve the Inequality

Use a graph to determine for which values of \( x \) the inequality \( \sqrt{2} \cos 2x \leq -1 \) holds. Since the cosine function oscillates between -1 and 1 and the inequality is non-strict (\( \leq \)), the regions where the inequality holds will be where the curve stays at or below \( -1 \). These are exactly the points where \( \cos 2x = -\frac{\sqrt{2}}{2} \), which occur at the same values previously calculated: \( x = \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8} \), when extended and evaluated graphically to confirm that the inequality holds for no other regions as the cosine doesn't sustain being \( -1 \) for more than instances.

Key Concepts

Cosine Function

The cosine function, expressed usually as \(\cos(x)\), is a fundamental part of trigonometry. It relates the angle of a right triangle to the ratio of the adjacent side over the hypotenuse. This periodic function oscillates between -1 and 1.
Its graph is a smooth wave that repeats every \(2\pi\), making it a periodic function essential for modeling cyclical behaviors in various fields, such as physics and engineering.
The transformation \(\cos(2x)\) indicates a change in frequency, causing the cycle to complete twice within the interval from 0 to \(2\pi\), effectively compressing the wave.Key elements to note about

• Maximum value: 1
• Minimum value: -1
• Period: \(2\pi\)
• Properties: Even function (\(\cos(-x) = \cos(x)\))

Understanding this helps in identifying the solutions to equations involving cosine by recognizing common angles and their corresponding cosine values.

Graphical Solution

A graphical solution offers a visual way to solve equations and inequalities. By plotting \(\sqrt{2} \cos(2x)\), we observe how the function behaves across the specified domain. This visual cue helps to determine where the function meets certain conditions like intersections or certain function values.
For solving the inequality \(\sqrt{2} \cos(2x) \leq -1\), sketching the graph gives an intuitive understanding of where the function values fall below or are equal to this threshold. It simplifies identifying solution intervals accurately.
Observing a graph can instantly reveal multiple solutions at different parts of a cycle, by indicating where the oscillations occur in relation to horizontal lines representing certain values. Hence, using graphs can speed up the solving process.Key points include:

• Checking where curves intersect specific values like \(-1\)
• Identifying different cycles and their effects on solutions
• Visualizing transformations caused by factors like \(2x\)

Thus, it provides a concrete representation of abstract mathematical concepts.

Inequalities

Inequalities describe a range of values rather than exact numbers. The inequality \(\sqrt{2} \cos(2x) \leq -1\) specifies that the transformed cosine function must be equal to or less than \(-1\). This shifts the focus from finding specific solutions to identifying all input values (\(x\)) that satisfy this condition.
The non-strict \(\leq\) indicates that the boundary value \(-1\) is included in potential solutions.
Meanwhile, inequalities like this assess when the potential shifts reach or overstep a specific value within a defined domain. Here's how it impacts solving:

• Identify points of equality as starting points
• Evaluate the behavior around these by comparing where the cosine values naturally fall in its cycle
• Graphically determine additional solution parts where these criteria apply

Ultimately, knowing how to interpret inequalities within trigonometric functions expands your toolkit to solve them effectively.

Interval Notation

Interval notation provides a concise method to describe continuous sets of numbers, commonly used alongside inequalities to express solution sets.
It uses brackets and parentheses to define the span of values included (or excluded) in the solutions, adopting this style:

• \([a, b]\) includes both endpoints \(a\) and \(b\)
• \((a, b)\) excludes endpoints
• \([a, b)\) or \((a, b]\) mix inclusion with exclusion

For our solutions, points like \(x = \frac{3\pi}{8}\) are pinpointed within an interval like \([0, 2\pi)\). The open endpoint at \(2\pi\) accentuates that this boundary isn't part of the solutions, which is particularly crucial in cyclical contexts like trigonometry.
Learning this notation helps elegantly communicate which portions satisfy equations or inequalities, supporting mathematical discussions and more complex evaluations.