Problem 55 - Precalculus: Functions and Graphs

Question

Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$2 \tan t-\sec ^{2} t=0$$

Step-by-Step Solution

Answer

Solutions are at \( t = \frac{\pi}{4} \) and \( t = \frac{5\pi}{4} \).

Step 1: Simplify the Trigonometric Equation

Start with the given equation: \[2 \tan t - \sec^{2} t = 0\]Rearrange terms to better analyze the equation:\[2 \tan t = \sec^{2} t\]

Step 2: Use Trigonometric Identity

Recall the trigonometric identity relating tangent and secant: \( \sec^2 t = 1 + \tan^2 t \). Substitute this into the equation:\[2 \tan t = 1 + \tan^2 t\]

Step 3: Form a Quadratic Equation

Rearrange the equation to form a standard quadratic equation in terms of \( \tan t \):\[ \tan^2 t - 2 \tan t + 1 = 0 \]

Step 4: Solve the Quadratic Equation

This equation is a perfect square trinomial. It can be factored as:\[ (\tan t - 1)^2 = 0 \]Solve for \( \tan t \):\[ \tan t = 1 \]

Step 5: Find Solutions Within the Interval

Find the values of \( t \) for which \( \tan t = 1 \) in the interval \([0, 2\pi)\). The tangent function equals 1 at \( \frac{\pi}{4} \) and again at \( \frac{5\pi}{4} \).

Key Concepts

Trigonometric Identities

Trigonometric identities are mathematical equations relating the trigonometric functions to each other. They are crucial for simplifying expressions and solving equations involving trigonometric functions. In this exercise, the identity \( \sec^2 t = 1 + \tan^2 t \) plays a fundamental role. This identity helps transform the given equation into a quadratic form.
Here’s why this identity is useful:

• It directly relates \( \sec^2 t \) and \( \tan^2 t \), allowing replacement to simplify the equation.
• By transforming the equation, it facilitates finding solutions in terms of a single trigonometric function, tangent in this case.

Regularly using such identities is beneficial; they simplify the process of solving complex trigonometric equations.

Tangent Function

The tangent function, represented as \( \tan(t) \), is one of the primary trigonometric functions. It is notable for its recurrence every \( \pi \) radians, making it periodic over its domain. To grasp it well:

• \( \tan(t) \) is the ratio of the sine and cosine functions: \( \tan(t) = \frac{\sin(t)}{\cos(t)} \).
• It is undefined where \( \cos(t) = 0 \), thus at odd multiples of \( \frac{\pi}{2} \).
• The tangent function equals 1 at angles such as \( \frac{\pi}{4} \) and repeats this value at subsequent intervals of \( \pi \).

In this problem, recognizing that \( \tan(t) = 1 \) at specific intervals helps determine the values of \( t \) that solve the equation within the specified interval.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, typically in the form of \( ax^2 + bx + c = 0 \). Solving quadratic equations is a fundamental skill in mathematics.
In our exercise, converting the problem into a quadratic form allowed us to use factoring to find solutions for \( \tan t \). Let's break down the process:

• First, rearrange the identity-based equation as \( \tan^2 t - 2 \tan t + 1 = 0 \), which fits the quadratic format.
• This particular equation is a perfect square trinomial, which simplifies to \( (\tan t - 1)^2 = 0 \).
• Solving the factorized form yields \( \tan t = 1 \).

Factoring when possible simplifies the computation and is a powerful method for finding solutions.

Solutions Over Interval

When solving trigonometric equations, determining solutions over a specific interval is often necessary. For this exercise, solutions needed to fit within \([0, 2\pi)\). Here's how to find these solutions:

• The interval \([0, 2\pi)\) defines a complete cycle for functions like tangent, covering all possible angle values before repeating.
• For \( \tan t = 1 \), the solutions initially occur at \( \frac{\pi}{4} \), and then at \( \frac{\pi}{4} + n\pi \) due to the tangent's periodicity, where \( n \) is an integer.
• Within \([0, 2\pi)\): identify values where \( n \) yields \( t \) within the interval, such as \( \frac{5\pi}{4} \).

A firm grasp of intervals and periodicity aids in quickly identifying valid solutions.