Problem 44 - Precalculus Mathematics for Calculus

Question

Solve the given equation. $$\tan ^{4} \theta-13 \tan ^{2} \theta+36=0$$

Step-by-Step Solution

Answer

The equation solutions are \( \theta = \tan^{-1}(3) + n\pi \), \( \theta = \tan^{-1}(-3) + n\pi \), \( \theta = \tan^{-1}(2) + n\pi \), and \( \theta = \tan^{-1}(-2) + n\pi \), \( n \in \mathbb{Z} \).

Step 1: Recognize Substitution Possibility

Notice that the equation \( \tan^{4} \theta - 13 \tan^{2} \theta + 36 = 0 \) is a quadratic in form if we use substitution. Let \( x = \tan^2 \theta \). This transforms the equation to \( x^2 - 13x + 36 = 0 \).

Step 2: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the values of \( x \). Here, \( a = 1 \), \( b = -13 \), and \( c = 36 \). Substitute these values:\[ x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4 \cdot 1 \cdot 36}}{2 \cdot 1} = \frac{13 \pm \sqrt{169 - 144}}{2} = \frac{13 \pm \sqrt{25}}{2} \]This gives us \( x = \frac{13 \pm 5}{2} \).

Step 3: Calculate Quadratic Roots

Calculate the roots: - \( x = \frac{13 + 5}{2} = 9 \)- \( x = \frac{13 - 5}{2} = 4 \)So, the solutions are \( x = 9 \) and \( x = 4 \).

Step 4: Retransform to Original Variable

Since \( x = \tan^2 \theta \), we have: - \( \tan^2 \theta = 9 \) implies \( \tan \theta = \pm 3 \)- \( \tan^2 \theta = 4 \) implies \( \tan \theta = \pm 2 \)

Step 5: General Solution

For \( \tan \theta = 3 \), the general solutions are \( \theta = \tan^{-1} (3) + n\pi \) and \( \theta = \tan^{-1} (-3) + n\pi \). For \( \tan \theta = 2 \), the general solutions are \( \theta = \tan^{-1} (2) + n\pi \) and \( \theta = \tan^{-1} (-2) + n\pi \) where \( n \in \mathbb{Z} \).

Key Concepts

Quadratic Substitution

Quadratic substitution is a powerful technique that helps transform complex equations into simpler forms. In the given problem, we start with the equation \( \tan^{4} \theta - 13 \tan^{2} \theta + 36 = 0 \). At first glance, solving this directly seems challenging due to terms like \( \tan^{4} \theta \). However, recognizing this equation's similarity to a quadratic allows us to make a substitution.

• Identify the quadratic-like structure: Here, the equation uses powers of \( \tan \theta \).
• Substitute \( x = \tan^2 \theta \), changing the trigonometric equation into a simpler quadratic one, \( x^2 - 13x + 36 = 0 \). This is now a standard quadratic equation that we can solve using familiar algebraic techniques.
• Quadratic substitution often uncovers hidden symmetry or simplification in trigonometric equations.

By transforming the trigonometric terms, we can apply straightforward algebra to advance the solution process, making it more approachable.

Tangent Function

The tangent function, \( \tan \theta \), is a fundamental concept in trigonometry that links an angle in a right triangle to the lengths of two of its sides. In a right triangle:

• \( \tan \theta = \frac{\text{opposite side}}{\text{adjacent side}} \)
• It has a period of \( \pi \), meaning \( \tan (\theta + n\pi) = \tan \theta \) for any integer \( n \).
• Unlike other basic trigonometric functions, it can take any real value, which makes it unique.

In this exercise, the tangent function appears as \( \tan^2 \theta \) and \( \tan^4 \theta \), which can be handled like any algebraic equations once a substitution is made. The properties of the tangent function, especially its infinite range, allow it to model periodic and continuous behavior in many real-world applications as well.

General Solution

A general solution in the context of trigonometric equations provides a method to capture all possible solutions for an angle \( \theta \). Trigonometric functions inherently repeat their values, thanks to their periodic nature, making their solutions extend into infinite sequences with common differences. Consider when \( \tan \theta = 3 \).

• One principal angle could be \( \theta = \tan^{-1}(3) \).
• However, due to the periodicity of the tangent function, a complete set of solutions must include repeats every \( \pi \) radians. Thus, the full general solution can be written as \( \theta = \tan^{-1}(3) + n\pi \), where \( n \) is any integer.

This approach applies equally to scenarios where \( \tan \theta = 2 \) or any other value. General solutions ensure every possibility is accounted for, crucial for comprehensive understanding and application.

Inverse Trigonometric Functions

Inverse trigonometric functions are used to find angle measures when given the ratios of the sides of a right triangle. The function \( \tan^{-1}(x) \) serves as the inverse of the tangent function, determining the angle \( \theta \) whose tangent is \( x \). Some key points include:

• \( \tan^{-1} \) provides principal values, typically restricted between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), ensuring a unique angle.
• When solving equations like \( \tan \theta = 3 \), \( \tan^{-1}(3) \) helps find one specific solution.
• Inverse functions are essential because they allow us to translate from ratio back to angle, a critical step in resolving many trigonometric problems.

They work hand in hand with periodicity to describe the full set of solutions for trigonometric equations, offering both precise values and the means to express infinite series of solutions.