Problem 81 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use inverse functions where necessary to solve the equation. $$\tan ^{2} x+\tan x-12=0$$

Step-by-Step Solution

Answer

The solutions of the given equation \(\tan ^{2} x+\tan x-12=0\) are \( x = -1.3258177...\) and \( x = 1.2490457...\)

Step 1: Identify the Quadratic Equation

The given equation is \(\tan ^{2} x+\tan x-12=0\). Let \(y = \tan{x}\). The equation then becomes a standard quadratic equation \( y^2 + y - 12 = 0 \)

Step 2: Factor the Quadratic Equation

Next, split the 'y' term such that the multiplication of two parts equals the product of 'a' and 'c' (here both are 1 and -12 respectively) and the addition equals 'b', which is 1. The factors of -12 which add up to 1 are 4 and -3. Hence, the equation is factored as \( (y + 4) (y - 3) = 0 \)

Step 3: Solve for y

Setting each factor in the equation equal to zero gives the solutions of 'y': \( y+4 = 0 => y = -4 \) and \( y-3 = 0 => y = 3 \)

Step 4: Substitute y with tan x

From step 3, we have two solutions for \(y\). But, we have substituted \(y = \tan{x}\) at the beginning. Substituting back, we have \( \tan{x} = -4 \) and \( \tan{x} = 3 \)

Step 5: Solve for x (Find Inverse Tangent)

Applying the inverse tangent to both sides of the equations, we get: \( x = \text{atan}(-4) = -1.3258177...\) and \( x = \text{atan}(3) = 1.2490457...\)

Key Concepts

Quadratic Equation

Quadratic equations are a fundamental element in algebra. A quadratic equation is typically expressed in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. The highest power of \( x \) is 2, indicating the equation is quadratic in nature. In the exercise, the quadratic equation appears as \( \tan^2 x + \tan x - 12 = 0 \), with \( y = \tan x \) being substituted to simplify it into \( y^2 + y - 12 = 0 \).

Solving quadratic equations involves processes like factoring, using the quadratic formula, or completing the square. In this instance, factoring was used to express the equation as \((y + 4)(y - 3) = 0\). This leads to solving two simpler linear equations: \( y + 4 = 0 \) and \( y - 3 = 0 \), yielding \( y = -4 \) and \( y = 3 \). These solutions set the stage for further exploration when substituting back \( \tan x \).

Trigonometric Functions

Trigonometric functions relate the angles and sides of a triangle to the unit circle. The primary trigonometric functions are sine, cosine, and tangent. In this exercise, tangent is the main function in use, denoted as \( \tan x \). Tangent calculates the ratio of the opposite side to the adjacent side of a right-angled triangle.

Understanding these functions is crucial in solving many mathematical problems, especially those involving periodic phenomena. In this problem, \( \tan x \) was squared and integrated into a quadratic form, demonstrating its flexibility in mathematical applications. As seen, transforming \( \tan x \) into \( y \) allowed the equation to be addressed via familiar quadratic solving techniques.

Since \( \tan x = -4 \) or \( \tan x = 3 \) involves angles whose tangent values are not standard, it exemplifies the need for inverse functions to find angles precisely.

Inverse Trigonometric Functions

Inverse trigonometric functions are crucial when you need to determine an angle given a trigonometric value. They are essentially the "reverse" functions of the basic trigonometric functions, allowing you to work backward from the value back to the angle. The inverse tangent, often written as \( \tan^{-1} \) or \( \text{atan} \), is the function utilized in finding the angle, given that \( \tan x = y \).

In our problem, after solving for \( y \), we revert to finding \( x \) by using the inverse tangent function. \( x = \text{atan}(-4) \) gives an angle of about \(-1.3258177\), while \( x = \text{atan}(3) \) results in approximately \(1.2490457\). These solutions show the importance of inverse functions in providing specific angle measurements, essential in fields like engineering and physics.

These functions are also integral for extending trigonometry into more complex applications, such as calculus, proving how vital it is to understand how they work in connection with standard trigonometric functions.