Problem 85 - Precalculus: Functions and Graphs

Question

Approximate the solution to each inequality on the interval \([0,2 \pi]\). $$\cos 3 x<\sin x$$

Step-by-Step Solution

Answer

The solutions on \([0, 2\pi]\) are \((0, \arctan(3)) \cup (\pi + \arctan(3), 2\pi)\).

Step 1: Set the inequality equation

Begin by setting the inequality equation. Given: \( \cos 3x < \sin x \) on the interval \([0, 2\pi]\). You need to find values of \(x\) where \( \cos 3x \) is less than \( \sin x \) within this domain.

Step 2: Find the critical points

Find points where \( \cos 3x = \sin x \). At these points, \( \cos 3x - \sin x = 0 \). Use the identity: \( \cos 3x - \sin x = 0 \) implies \( \tan x = 3 \). Calculate \( x = \arctan(3) \). This gives us one critical point.

Step 3: Solve within the range

You need to account for the periodicity of trigonometric functions. The equation \( \tan x = 3 \) will be satisfied each time \( x + n\pi = \arctan(3) \), where \( n \) is an integer. Solve for \( x \) within \([0, 2\pi]\): locating \( x_1 = \arctan(3) \approx 1.249 \) and, by symmetry, \( x_2 = \pi + \arctan(3) \approx 4.390 \).

Step 4: Analyze intervals

Analyze points between key turning points within the interval \([0, 2\pi]\). Evaluate \( \cos 3x \) and \( \sin x \) between \( 0 \) and \( x_1 \), between \( x_1 \) and \( x_2 \), and between \( x_2 \) and \( 2\pi \). The inequality holds when \( \cos 3x < \sin x \).

Step 5: Determine solution intervals

Determine where \( \cos 3x < \sin x \) by testing values in each interval from Step 4. You find that the inequality holds in intervals:\( (0, \arctan(3)) \) and \( (\pi + \arctan(3), 2\pi) \).

Step 6: Write the solution

Conclude that the solutions to \( \cos 3x < \sin x \) over \([0, 2\pi]\) are \(x \in (0, \arctan(3)) \cup (\pi + \arctan(3), 2\pi) \).

Key Concepts

Critical Points

In trigonometric inequalities, critical points are where two trigonometric functions intersect. For the inequality \( \cos 3x < \sin x \), finding the critical points starts with setting the equation \( \cos 3x = \sin x \). At these points, the functions are equal, effectively "balancing" each other out.

This is important because these points mark potential boundaries where the inequality may change from true to false, or vice versa. Calculating critical points involves using trigonometric identities or transformations. Here, \( \cos 3x - \sin x = 0 \) is simplified to \( \tan x = 3 \). The critical point can then be found using the inverse tangent: \( x = \arctan(3) \).

Finding critical points is foundational as they determine where solutions to inequalities may start or end within a given interval.

Periodicity of Trigonometric Functions

The periodicity of trigonometric functions describes how functions repeat themselves over regular intervals. Key trigonometric functions like sine and cosine have intrinsic periods—such as \(2\pi\) for \(\sin(x)\) and \(\cos(x)\), meaning they return to the same value every \(2\pi\) radians.

In our problem, the periodic nature of \(\tan(x)\) impacts the solution. When \( \tan x = 3 \), the condition \( x + n\pi = \arctan(3) \) holds, with \( n \) being any integer. This repetition is due to the \(\pi\)-periodicity of the tangent function. Therefore, \(\tan(x)\) returns to the same value in additional intervals of \(\pi\).

This periodic property helps us locate all possible solutions within a specified range, ensuring we consider all such points in the solution set.

Solution Intervals

After determining critical points and understanding periodicity, we analyze where the inequality \( \cos 3x < \sin x \) is true within specified intervals. By setting up intervals like \([0, \arctan(3)]\) and \([\pi + \arctan(3), 2\pi]\), we place conditions to test where the inequality holds.

One examines the behavior of \( \cos 3x \) and \( \sin x \) within each segment by picking test points or using graphical analysis.

• Between \(0\) and \(x_1 = \arctan(3)\), \( \cos 3x \) is less than \( \sin x \).
• Between \(x_1\) and \(x_2 = \pi + \arctan(3)\), this condition may no longer be true, demonstrating a possible reversal in the inequality.
• Between \(x_2\) and \(2\pi\), the condition returns to true.

Hence, the intervals \((0, \arctan(3))\) and \((\pi + \arctan(3), 2\pi)\) satisfy the inequality, revealing where the solutions lie.