Problem 83 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use inverse functions where necessary to solve the equation. $$\sec ^{2} x-6 \tan x+4=0$$

Step-by-Step Solution

Answer

General solutions are \( x = \arctan (5) + n\pi \) and \( x = \arctan (1) + n\pi \), where \( n \) is an integer.

Step 1: Simplify the Equation

First, we simplify the equation by converting the secant function into tangent function. We use the identity \( \sec ^{2} x = 1 + \tan ^{2} x \). Therefore, our equation becomes \( 1 + \tan ^{2} x - 6 \tan x + 4 = 0 \). This can be rearranged into a quadratic equation in terms of \( \tan x \), i.e., \( \tan ^{2} x - 6 \tan x + 5 = 0 \).

Step 2: Solve the Quadratic Equation

We solve the above quadratic equation as any regular quadratic equation to get the values of \( \tan x \). Through factorization, the equation becomes: \( (\tan x - 5)(\tan x - 1) = 0 \). Hence, \( \tan x = 5 \) or \( \tan x = 1 \).

Step 3: Find the Values of \( x \)

Finally, find the values of \( x \) by applying the inverse tangent function to 5 and 1. \( x = \arctan (5) \) or \( x = \arctan (1) \). As the tangent function has a period of \( \pi \), the general solutions are \( x = \arctan (5) + n\pi \) and \( x = \arctan (1) + n\pi \), where \( n \) is an integer.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions help us find the angle that corresponds to a given trigonometric ratio. They're denoted as \(\sin^{-1}, \cos^{-1}, \tan^{-1}\), and so forth. In this problem, once you've found the values of \(\tan x\) from solving the quadratic equation, inverse trigonometric functions are used to determine the angles \(x\).
Imagine you know \(\tan x = 1\). By using the inverse tangent function, \(x = \arctan(1)\), which tells us the angle whose tangent is 1.
The angle \(x = \arctan(1)\) is \(\frac{\pi}{4}\), and because tangent is periodic (repeats every \(\pi\)), the general solution includes \(x = \arctan(1) + n\pi\), where \(n\) is any integer.
The inverse function helps identify these specific angle values from unlimited possibilities.

Quadratic Equations

Quadratic equations are polynomial equations that involve terms up to the second degree, typically written as \(ax^2 + bx + c = 0\). In this exercise, the equation transformed into a quadratic form once the trigonometric identities were applied: \(\tan^2 x - 6 \tan x + 5 = 0\).
To solve a standard quadratic equation, you can use:

• Factorization, as done here, if the equation can be neatly expressed as a product of two binomials."
• The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), applicable when simpler methods aren't evident.

In this case, factorization is efficient since the roots are straightforward: \((\tan x - 5)(\tan x - 1) = 0\) gives the solutions \(\tan x = 5\) and \(\tan x = 1\). These solutions are crucial for applying inverse functions to find actual angle values.

Trigonometric Identities

Trigonometric identities like \(\sec^2 x = 1 + \tan^2 x\) are essential tools in solving trigonometric equations. They allow converting complex forms into simpler or more familiar expressions.
In our specific problem, the identity \(\sec^2 x = 1 + \tan^2 x\) is used to rewrite \(\sec^2 x - 6 \tan x + 4 = 0\) into \(1 + \tan^2 x - 6 \tan x + 4 = 0\).
This substitution is key to forming a recognizable quadratic equation. Identifying the right identity can often bring a seemingly difficult problem into a realm where common algebraic techniques can be applied.
Mastering these identities opens up multiple strategies to approach trigonometric problems and analyze their components efficiently.