Problem 20 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation for solutions over the interval \([0,2 \pi)\) by first solving for the trigonometric finction. Do not use a calculator. $$\tan x-\cot x=0$$

Step-by-Step Solution

Answer

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

Step 1: Rewrite the Cotangent Function

Rewrite the equation \( \tan x - \cot x = 0 \) using the identity \( \cot x = \frac{1}{\tan x} \). This gives us the equation \( \tan x - \frac{1}{\tan x} = 0 \).

Step 2: Clear the Fraction

Multiply every term in the equation \( \tan x - \frac{1}{\tan x} = 0 \) by \( \tan x \) to eliminate the fraction: \( \tan^2 x - 1 = 0 \).

Step 3: Solve the Quadratic Equation

The equation \( \tan^2 x - 1 = 0 \) is quadratic in form. Factor it into \((\tan x - 1)(\tan x + 1) = 0 \). This gives us two possible equations: \( \tan x - 1 = 0 \) and \( \tan x + 1 = 0 \).

Step 4: Solve for \( \tan x = 1 \)

Solve \( \tan x = 1 \). The solutions are \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \) because tangent is positive in the first and third quadrants.

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

Solve \( \tan x = -1 \). The solutions are \( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \) because tangent is negative in the second and fourth quadrants.

Key Concepts

Tangent Function

The tangent function, denoted as \( \tan x \), is a fundamental trigonometric function that relates the angle of a right triangle to the ratios of two of its sides. Specifically, in a right triangle, the tangent of angle \( x \) is defined as the ratio of the length of the opposite side to the length of the adjacent side. For any angle \( x \), the tangent function can be expressed as:

• \( \tan x = \frac{\sin x}{\cos x} \)

This definition highlights two important aspects of the tangent function:

• It is periodic, with a period of \( \pi \), which means the pattern of the function repeats every \( \pi \) units.
• It is undefined when \( \cos x = 0 \) (i.e., \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots \) ), causing vertical asymptotes at these points.

The tangent function is crucial when solving equations like \( \tan x = 1 \) or \( \tan x = -1 \), as these equations help determine specific angles at which the ratios equal 1 or -1. Solutions are typically found within specified intervals, such as \( [0, 2\pi) \), accounting for all possible angles that satisfy the equation within a complete cycle of the unit circle.

Cotangent Function

The cotangent function, symbolized as \( \cot x \), is the reciprocal of the tangent function. This means that for any angle \( x \), the cotangent is the ratio of the adjacent side to the opposite side in a right triangle. Mathematically, it is defined as:

• \( \cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x} \)

The cotangent function has some distinctive characteristics:

• It is periodic with a period of \( \pi \), similar to the tangent function.
• It is undefined at points where \( \sin x = 0 \) (i.e., \( x = 0, \pi, 2\pi, \ldots \) ), resulting in vertical asymptotes at these angles.

When solving trigonometric equations involving cotangent, such as \( \tan x - \cot x = 0 \), it's essential to recognize the roles both tangent and cotangent play as reciprocal functions. Substituting \( \cot x = \frac{1}{\tan x} \) often facilitates the equation's simplification, making it easier to solve by transforming the expression into a quadratic form.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, commonly taking the form \( ax^2 + bx + c = 0 \). In the context of trigonometric equations, such as \( \tan^2 x - 1 = 0 \), they appear once we clear fractions or simplify expressions involving trigonometric identities. Breaking down a trigonometric equation into a quadratic form involves:

• Transforming terms using identities, such as \( \cot x = \frac{1}{\tan x} \).
• Eliminating fractions by multiplying through by denominators.
• Commanding algebraic techniques to factor and solve for roots.

To solve \( \tan^2 x - 1 = 0 \), we derive:

• Factor into \( (\tan x - 1)(\tan x + 1) = 0 \).
• Set each factor to zero: \( \tan x - 1 = 0 \) and \( \tan x + 1 = 0 \).
• Solve these to find \( x \) values that satisfy each equation within the interval \([0, 2\pi)\).

Understanding quadratic equations strengthens your ability to solve a wide array of mathematical problems, including those involving trigonometric functions.