Problem 45 - Precalculus: Functions and Graphs

Question

Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$\tan ^{2} x \sin x=\sin x$$

Step-by-Step Solution

Answer

The solutions are \( x = 0, \pi, \frac{\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4} \).

Step 1: Factor the Equation

Start with the given equation \( \tan^2 x \sin x = \sin x \) and notice that \( \sin x \) is a common factor on both sides of the equation. We can factor \( \sin x \) out of the equation to get \( \sin x (\tan^2 x - 1) = 0 \).

Step 2: Solve for \( \sin x = 0 \)

Set the first factor equal to zero: \( \sin x = 0 \). Solve for \( x \) in the interval \( [0, 2\pi) \). Thus, \( x = 0, \pi \).

Step 3: Solve for \( \tan^2 x - 1 = 0 \)

Set the second factor equal to zero: \( \tan^2 x - 1 = 0 \). This implies \( \tan^2 x = 1 \), which leads to \( \tan x = \pm 1 \).

Step 4: Find Solutions for \( \tan x = 1 \)

Solve \( \tan x = 1 \). The solutions are \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \) within \( [0, 2\pi) \).

Step 5: Find Solutions for \( \tan x = -1 \)

Solve \( \tan x = -1 \). The solutions are \( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \) within \( [0, 2\pi) \).

Key Concepts

Factoring Trigonometric Functions

When solving trigonometric equations, factoring is a powerful technique that helps simplify and solve complex expressions. In our exercise, we started with the equation \( \tan^2 x \sin x = \sin x \). Notice that \( \sin x \) is common on both sides. To factor, we extracted \( \sin x \), transforming our equation to \( \sin x (\tan^2 x - 1) = 0 \).
This step is crucial because it breaks down the original equation into simpler components. Each factor can now be set equal to zero, giving us two simpler equations to solve:

• \( \sin x = 0 \)
• \( \tan^2 x - 1 = 0 \)

By factoring, we reduce the problem to solving two equations, making it easier to find the solutions within the given interval.

Interval Notation

In mathematics, interval notation is a compact and efficient way to denote ranges of values, especially real numbers. The interval \([0, 2\pi)\) indicates all the real numbers starting from 0 up to, but not including, \(2\pi\).
This notation uses brackets:

• \([ ]\) means the endpoints are included, or 'closed interval'
• \(( )\) means the endpoints are not included, or 'open interval'

For our problem, the solutions must lie within the interval \([0, 2\pi)\). This means that the solutions we identify – either for \( \sin x = 0 \) or \( \tan^2 x - 1 = 0 \) – should fall within this range. Understanding how to read and interpret interval notation is key to correctly identifying all possible solutions.

Tangent Function

The tangent function, \( \tan x \), is one of the fundamental trigonometric functions and is defined as the ratio of the sine and cosine functions: \( \tan x = \frac{\sin x}{\cos x} \). Understanding how the tangent function behaves is important when dealing with equations involving tangent.
Note that \( \tan x \) has asymptotes, or values it cannot take, at \( x = \frac{\pi}{2}, \frac{3\pi}{2},\) etc., because there \( \cos x = 0 \). The tangent function has a period of \( \pi \), meaning it repeats every \( \pi \) units.

• When \( \tan x = 1 \), typically at \( x = \frac{\pi}{4} \) and repeated every \( \pi \) units
  
• When \( \tan x = -1 \), typically at \( x = \frac{3\pi}{4} \) and repeated every \( \pi \) units
  

In our problem, solving \( \tan^2 x = 1 \) gives us \( \tan x = \pm1 \), leading to specific solutions in the interval \([0, 2\pi)\). Knowing these properties helps you predict and verify the solutions accurately.