Problem 36 - Precalculus: Functions and Graphs

Question

Find all solutions of the equation. $$\tan \alpha+\tan ^{2} \alpha=0$$

Step-by-Step Solution

Answer

The solutions are \( \alpha = n\pi \) and \( \alpha = -\frac{\pi}{4} + n\pi \), where \( n \) is an integer.

Step 1: Start with the Original Equation

Consider the original equation: \( \tan \alpha + \tan^2 \alpha = 0 \). Our goal is to find all values of \( \alpha \) that satisfy this equation.

Step 2: Factorize the Equation

Notice that we can factor the equation as follows: \( \tan^2 \alpha + \tan \alpha = 0 \) can be rewritten as \( \tan \alpha (\tan \alpha + 1) = 0 \).

Step 3: Solve Each Factor Separately

Since the product is zero, use the zero-product property: either \( \tan \alpha = 0 \) or \( \tan \alpha + 1 = 0 \). This gives two scenarios to solve.

Step 4: Solve \( \tan \alpha = 0 \)

The tangent function is zero at integer multiples of \( \pi \), so \( \alpha = n\pi \), where \( n \) is any integer.

Step 5: Solve \( \tan \alpha + 1 = 0 \)

Rearrange to find \( \tan \alpha = -1 \). The tangent function is \(-1\) at odd multiples of \( \frac{\pi}{4} \): \( \alpha = -\frac{\pi}{4} + n\pi \), where \( n \) is any integer.

Step 6: Combine the Solutions

The general solutions are \( \alpha = n\pi \) and \( \alpha = -\frac{\pi}{4} + n\pi \), where \( n \) is an integer. In set notation, it is \( \alpha = \{ n\pi \} \cup \{ -\frac{\pi}{4} + n\pi \} \).

Key Concepts

Tangent Function

The tangent function, denoted as \( \tan \alpha \), is one of the primary trigonometric functions, alongside sine and cosine. It is defined as the ratio of the sine function to the cosine function: \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \] This function is periodic with a period of \( \pi \), meaning it repeats its values every \( \pi \) units along the \( \alpha \)-axis. The tan function has vertical asymptotes where the cosine function equals zero, occurring at odd multiples of \( \frac{\pi}{2} \).

• Values of \( \alpha \) at which \( \tan \alpha = 0 \) occur at integer multiples of \( \pi \).
• Tangent is positive in the first and third quadrants, while negative in the second and fourth.

Understanding these properties is essential in solving trigonometric equations involving the tangent function.

Zero-product property

The zero-product property is a fundamental principle in algebra, stating that if the product of two factors is zero, at least one of the factors must be zero. For the equation \( \tan \alpha + \tan^2 \alpha = 0 \), we can factor it into \( \tan \alpha (\tan \alpha + 1) = 0 \). Applying the zero-product property, this gives us two separate scenarios to consider:

• Case 1: \( \tan \alpha = 0 \)
• Case 2: \( \tan \alpha + 1 = 0 \)

By considering each factor individually, we can find more potential solutions to the equation. This approach simplifies complex expressions into more manageable parts.

Integer Multiples of Pi

The concept of integer multiples of \( \pi \) primarily arises in trigonometry due to the periodic nature of trigonometric functions. When we refer to integer multiples of \( \pi \), it involves values of \( \alpha \) such as \( 0, \pi, 2\pi, 3\pi, ... \), which effectively means \( \alpha = n\pi \) where \( n \) is an integer.

• For \( \tan \alpha = 0 \), solutions occur precisely at these integer multiples because \( \tan(n\pi) = 0 \) for any integer \( n \).
• This periodic repetition is due to the complete oscillation of sine and cosine, which defines tangent.

Understanding this periodicity helps in predicting and calculating specific angles that solve trigonometric equations.

General Solutions

General solutions in trigonometric equations encapsulate all possible solutions derived from the periodicity of trigonometric functions. In our given problem, the general solution for \( \tan \alpha = 0 \) is \( \alpha = n\pi \) while for \( \tan \alpha = -1 \), it is \( \alpha = -\frac{\pi}{4} + n\pi \), where \( n \) can be any integer.

• The first solution covers cases where the tangent is zero, and the second accounts for when it equals \(-1\).
• By including \( n\pi \), the periodic nature of the tangent function is considered, making the solution comprehensive.

General solutions are powerful because they offer a compact expression for infinite solutions, reflecting the inherent periodicity found in trigonometric equations.