Problem 50 - Precalculus Mathematics for Calculus

Question

Solve the given equation. $$3 \tan ^{3} \theta=\tan \theta$$

Step-by-Step Solution

Answer

\( \theta = n\pi, \frac{\pi}{6} + n\pi, \frac{5\pi}{6} + n\pi \) for integers \( n \).

Step 1: Analyze the Equation

The given equation is \( 3 \tan ^{3} \theta = \tan \theta \). We need to find the values of \( \theta \) where this equation holds true.

Step 2: Simplifying the Equation

Factor out \( \tan \theta \) on both sides to rewrite the equation: \( 3 \tan^3 \theta - \tan \theta = 0 \). This turns into \( \tan \theta (3 \tan^2 \theta - 1) = 0 \).

Step 3: Set Each Factor to Zero

The equation \( \tan \theta (3 \tan^2 \theta - 1) = 0 \) gives us two separate equations: \( \tan \theta = 0 \) and \( 3 \tan^2 \theta - 1 = 0 \).

Step 4: Solve for \( \tan \theta = 0 \)

For \( \tan \theta = 0 \), \( \theta = n\pi \), where \( n \) is any integer, because \( \tan \theta \) equals zero at integer multiples of \( \pi \).

Step 5: Solve for \( 3 \tan^2 \theta - 1 = 0 \)

For the equation \( 3 \tan^2 \theta - 1 = 0 \), rearrange to get \( 3 \tan^2 \theta = 1 \), which simplifies to \( \tan^2 \theta = \frac{1}{3} \). This implies \( \tan \theta = \pm \frac{1}{\sqrt{3}} \).

Step 6: Solve for Specific \( \theta \) values

The values \( \tan \theta = \frac{1}{\sqrt{3}} \) and \( \tan \theta = -\frac{1}{\sqrt{3}} \) correspond to angles \( \theta = \frac{\pi}{6} + n\pi \) and \( \theta = \frac{5\pi}{6} + n\pi \) respectively, where \( n \) is an integer because tangent has a period of \( \pi \).

Step 7: Combine the Solutions

Combining the solutions from Steps 4 and 6, the solution is \( \theta = n\pi \), \( \frac{\pi}{6} + n\pi \), and \( \frac{5\pi}{6} + n\pi \) for any integer \( n \).

Key Concepts

Tangent Function

The tangent function, often denoted as \( \tan \theta \), is a fundamental trigonometric function. It is defined as the ratio of the sine to the cosine of an angle, that is, \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). This function is well-known for its unique characteristic of having no limits when approaching vertical asymptotes, occurring at angles where \( \cos \theta = 0 \).

• The tangent function is useful in various trigonometric equations and problems.
• It exhibits particular behaviors over different intervals, making it a favorite in describing periodic phenomena.

Understanding how to manipulate and solve equations involving \( \tan \theta \) is crucial, as shown in the original problem where factoring \( \tan \theta \) helped reduce the equation to simpler parts for further solution.

Periodic Functions

Periodic functions repeat their values at regular intervals. The tangent function is one of the basic examples of such a function.

• The period of the tangent function is \( \pi \), meaning its values repeat every \( \pi \) radians.
• This periodicity simplifies solving trigonometric equations, as solutions often include an integer multiple of the period.

For example, in the equation \( \tan \theta = 0 \), solutions were found at \( \theta = n\pi \) because \( \tan \theta \) equals zero at these integer multiples of \( \pi \). Periodicity is key when considering the full set of solutions in trigonometric problems.

Factoring Equations

Factoring is a mathematical process critical to simplifying and solving equations. In trigonometric equations, factoring can separate complex expressions into manageable parts.

• The original exercise demonstrated factoring by expressing \( 3 \tan^3 \theta - \tan \theta = 0 \) as \( \tan \theta (3 \tan^2 \theta - 1) = 0 \).
• By making each factor zero, separate and simpler equations arise, which are more straightforward to solve.

These techniques help break down potentially complex trigonometric problems, making discovering all possible solutions achievable.

Trigonometric Solutions

Solving trigonometric equations involves determining the angle \( \theta \) that satisfies the equation for all applicable cases.

• In our example, solutions for \( \tan \theta = 0 \) were found at \( \theta = n\pi \), a straightforward result due to the periodicity of the tangent function.
• The additional solutions for \( \tan \theta = \pm \frac{1}{\sqrt{3}} \) introduced sequences \( \theta = \frac{\pi}{6} + n\pi \) and \( \theta = \frac{5\pi}{6} + n\pi \), derived from specific unit circle values.

Combining these solutions encompasses all possible values of \( \theta \) that make the original equation valid. Each solution includes an integer \( n \), ensuring coverage over all cycles of the periodic function.