Problem 86 - Precalculus: Functions and Graphs

Question

Approximate the solution to each inequality on the interval \([0,2 \pi]\). $$\tan x \leq \sin 2 x$$

Step-by-Step Solution

Answer

Solutions are \( x = \pi, \frac{5\pi}{4}, \frac{7\pi}{4} \).

Step 1: Use a Trigonometric Identity

The given inequality is \( \tan x \leq \sin 2x \). Recall that \( \tan x = \frac{\sin x}{\cos x} \) and \( \sin 2x = 2 \sin x \cos x \). We substitute these identities into the inequality, obtaining \( \frac{\sin x}{\cos x} \leq 2 \sin x \cos x \).

Step 2: Clear the Fraction

Multiply both sides of the inequality by \( \cos x \) (assuming \( \cos x eq 0 \) to avoid division by zero), resulting in \( \sin x \leq 2 \sin x \cos^2 x \).

Step 3: Rearrange the Inequality

Rearrange the inequality to isolate \( \sin x \) terms on one side: \( \sin x (1 - 2 \cos^2 x) \leq 0 \).

Step 4: Factor the Expression

Recognize that \( 1 - 2 \cos^2 x = - (2 \cos^2 x - 1) \). Set each factor less than or equal to zero: \( \sin x \leq 0 \) and \( 2 \cos^2 x - 1 \geq 0 \).

Step 5: Solve Each Factor Inequality - for \( \sin x \leq 0 \)

The inequality \( \sin x \leq 0 \) implies that \( x \) is in intervals where the sine function is non-positive: \([\pi, 2\pi]\).

Step 6: Solve Each Factor Inequality - for \( 2 \cos^2 x - 1 \geq 0 \)

Rearrange to \( \cos^2 x \geq \frac{1}{2} \), meaning \( \cos x \geq \frac{1}{\sqrt{2}} \) or \( \cos x \leq -\frac{1}{\sqrt{2}} \). Solve for \( x \) between \( [0, 2\pi] \).

Step 7: Determine Intersections and Approximate Solutions

Combine solutions from Steps 5 and 6 to find common intervals. Check intersections, yielding \( x = \pi, \frac{5\pi}{4}, \frac{7\pi}{4} \).

Step 8: Verify and Conclude

Verify each solution by substitution into the original inequality. Confirm \( x = \pi, \frac{5\pi}{4}, \frac{7\pi}{4} \) satisfy \( \tan x \leq \sin 2 x \).

Key Concepts

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for every value of the variable where both sides are defined. In our problem, we used two important trigonometric identities:

• \( \tan x = \frac{\sin x}{\cos x} \)
• \( \sin 2x = 2 \sin x \cos x \)

These identities allow us to transform the original inequality \( \tan x \leq \sin 2x \) into an expression that is easier to work with: \( \frac{\sin x}{\cos x} \leq 2 \sin x \cos x \). By substituting the identities, we break down complex expressions into more manageable parts, enabling us to solve the inequality effectively.
This step of substitution is crucial as it utilizes known relationships between trigonometric functions to simplify the inequality and make the problem-solving process more straightforward.

Unit Circle

The unit circle is a fundamental concept in trigonometry that facilitates understanding of the behavior of trigonometric functions. It is a circle with a radius of 1, centered at the origin of a coordinate plane.
On the unit circle:

• The \( x \)-coordinate of a point represents the cosine of the angle.
• The \( y \)-coordinate represents the sine of the angle.

In our solution, when solving inequalities such as \( \sin x \leq 0 \), the unit circle helps identify the range of angles where the sine value is non-positive, which are angles from \( \pi \) to \( 2\pi \).
Similarly, it aids in visualizing and solving \( \cos^2 x \geq \frac{1}{2} \) by identifying the angles where the cosine value fulfills the condition. Using the unit circle is especially valuable for understanding sign changes and periodic properties of trigonometric functions.

Inequality Solving

Solving inequalities involves finding the set of values that satisfy the inequality condition. In the context of trigonometric inequalities, this process can seem complex but can be broken down step by step.
Initially, we need to simplify and rearrange the inequality using identities and algebraic manipulation. In our example, \( \sin x (1 - 2 \cos^2 x) \leq 0 \) emerges after clearing fractions and rearranging terms.

• Factorization is then used to break it into simpler components: \( \sin x \leq 0 \) and \( 2 \cos^2 x - 1 \geq 0 \).
• Each component inequality is solved separately, which is simpler to handle.

This methodical approach allows us to understand and solve inequalities by analyzing one factor at a time, before combining solutions for final results.

Intersection of Intervals

In solving trigonometric inequalities, finding the intersection of intervals is key to combining solutions from different conditions. It is the process of identifying common values that satisfy all parts of the inequality.
After finding solutions to each part of our factorized inequality, we need to determine where these solutions overlap.

• The intervals for \( \sin x \leq 0 \) were \([\pi, 2\pi]\).
• The intervals for \( \cos^2 x \geq \frac{1}{2} \) were derived from \( \cos x \geq \frac{1}{\sqrt{2}} \) or \( \cos x \leq -\frac{1}{\sqrt{2}} \).

The common intervals are determined by mapping solutions and identifying overlaps, yielding values where both inequalities are satisfied, like \( x = \pi, \frac{5\pi}{4}, \frac{7\pi}{4} \).
This method is crucial as it ensures all conditions of the problem are met, leading to a complete and correct solution.