Problem 32 - A Graphical Approach to Precalculus with Limits

Question

Solve \((\mathbf{a}) f(x)=0,(\mathbf{b}) f(x)>0,\) and \((\mathbf{c}) f(x)<0\) over the interval \([0,2 \pi)\) $$f(x)=2 \tan ^{2} x \sin x-\tan ^{2} x$$

Step-by-Step Solution

Answer

(a) \( x = 0, \frac{\pi}{6}, \pi, \frac{5\pi}{6} \); (b) \( x \in \left(\frac{\pi}{6}, \frac{5\pi}{6}\right) \); (c) \( x \in (0, \frac{\pi}{6}) \cup \left(\frac{5\pi}{6}, 2\pi\right) \).

Step 1: Simplify the Function

Start with the function given: \( f(x) = 2 \tan^2 x \sin x - \tan^2 x \). Factor out \( \tan^2 x \) to simplify the expression:\[f(x) = \tan^2 x (2 \sin x - 1)\]

Step 2: Solve Part (a) - Find where \( f(x) = 0 \)

Since \( f(x) = \tan^2 x (2 \sin x - 1) = 0 \), set each factor to zero separately:1. \( \tan^2 x = 0 \): This implies \( \tan x = 0 \), which occurs when \( x = n\pi \) for integer \( n \). Within \([0, 2\pi)\), \( x = 0 \) and \( x = \pi \).2. \( 2 \sin x - 1 = 0 \): Solve \( \sin x = \frac{1}{2} \). This occurs at \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).Thus, the solutions for \( f(x) = 0 \) are \( x = 0, \frac{\pi}{6}, \pi, \frac{5\pi}{6} \).

Step 3: Solve Part (b) - Find where \( f(x) > 0 \)

Since the expression is \( \tan^2 x (2 \sin x - 1) > 0 \), consider two cases:- \( \tan^2 x > 0 \) and \( 2 \sin x - 1 > 0 \) occurs when \( \sin x > \frac{1}{2} \), so \( x \in \left(\frac{\pi}{6}, \frac{5\pi}{6}\right)\).- \( \tan^2 x < 0 \) never happens as \( \tan^2 x \) is non-negative, so this situation does not contribute.Thus, \( f(x) > 0 \) in the intervals \( x \in \left(\frac{\pi}{6}, \frac{5\pi}{6}\right)\).

Step 4: Solve Part (c) - Find where \( f(x) < 0 \)

Now, consider when \( \tan^2 x (2 \sin x - 1) < 0 \). This can only occur when one factor is positive and the other negative:- \( \tan^2 x > 0 \) and \( 2 \sin x - 1 < 0 \) occurs when \( \sin x < \frac{1}{2} \), so \( x \in (0, \frac{\pi}{6}) \cup \left(\frac{5\pi}{6}, \pi\right) \cup \left(\pi, 2\pi\right) \).Thus, \( f(x) < 0 \) in the intervals \( x \in (0, \frac{\pi}{6}) \cup \left(\frac{5\pi}{6}, 2\pi\right) \).

Key Concepts

Tangent Function

The tangent function, often denoted as \( \tan x \), is an essential function in trigonometry. It is derived from the sine and cosine functions, specifically via the equation \( \tan x = \frac{\sin x}{\cos x} \). This relation highlights that tangent is the ratio of the opposite side to the adjacent side in a right-angled triangle.

Key characteristics of the tangent function include:

• Periodicity: The tangent function has a period of \( \pi \), meaning that it repeats its values every \( \pi \) units.
• Undefined Points: It becomes undefined at odd multiples of \( \frac{\pi}{2} \) because the cosine function is zero at these points, leading to a division by zero.
• Behavior: Unlike sine and cosine, which oscillate between -1 and 1, tangent can take any real number value. It also exhibits vertical asymptotes where it becomes undefined.

In solving equations like \( f(x) = 2 \tan^2 x \sin x - \tan^2 x \), identifying where \( \tan^2 x \) equals zero is crucial. Since \( \tan^2 x \) is always non-negative, it simplifies many problems by allowing us to focus on the zero-points of tangent itself.

Sine Function

The sine function, denoted as \( \sin x \), plays a vital role in trigonometry alongside the cosine function. Its fundamental attributes make it a staple in analyzing trigonometric equations.

Key aspects of the sine function include:

• Periodicity: The sine function has a period of \( 2\pi \), repeating every \( 2\pi \) units.
• Range: Its values oscillate between -1 and 1.
• Zeros: Sine is zero at integer multiples of \( \pi \), specifically 0, \( \pi \), and \( 2\pi \) within the interval \([0, 2\pi)\).

When solving inequalities or set equations like \( 2 \sin x - 1 = 0 \), recalling the values for which sine equals \( \frac{1}{2} \) is beneficial. Specifically, \( \sin x = \frac{1}{2} \) at \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \), making these values critical checkpoints while examining functions involving sine.

Inequalities

Trigonometric inequalities involve expressions where trigonometric functions are compared to constants or other functions. Solving these requires a strategic approach to evaluate the conditions leading to the inequality's truth.

Important steps when solving inequalities like \( f(x) > 0 \) or \( f(x) < 0 \) include:

• Factorization: As seen in the given exercise, factorizing the function helps identify when parts like \( \tan^2 x \) or \( 2 \sin x - 1 \) influence the inequality.
• Test Intervals: For functions over an interval like \([0,2\pi)\), testing sub-intervals—where function factors turn zero or change sign—is crucial.
• Combine Conditions: Analyze each factor distinctly but ensure you understand their combined effect on the inequality.

For the function \( f(x) = 2 \tan^2 x \sin x - \tan^2 x \), you rely on understanding where both \( \tan^2 x \) and \( \sin x \) either make the whole function positive or negative, guiding you to pertinent sections of the angle interval.