Problem 54 - Algebra and Trigonometry

Question

solve each equation on the interval \([0,2 \pi)\) $$ (\tan x+1)(\sin x-1)=0 $$

Step-by-Step Solution

Answer

The solutions to the equation \((\tan x+1)(\sin x-1)=0\) in the interval \(0 \leq x < 2\pi\) are \(x=\frac{\pi}{2}, \frac{3\pi}{4}, \frac{7\pi}{4}\).

Step 1: Set the first expression equal to zero

Set \(\tan x + 1 = 0\) and solve for x. This gives \(\tan x = -1\). The solutions to this equation within the specified interval are \(x=\frac{3\pi}{4}, \frac{7\pi}{4}\).

Step 2: Set the second expression equal to zero

Set \(\sin x - 1 = 0\) and solve for x. This gives \(\sin x = 1\). The solution to this equation within the specified interval is \(x=\frac{\pi}{2}\).

Step 3: Combine the solutions

The solutions to the equation are the unique combined solutions to the two expressions, within the interval \(0 \leq x < 2\pi\).

Key Concepts

Tangent Function

The tangent function, often abbreviated as "tan," is one of the fundamental trigonometric functions. It is defined as the ratio of the sine function to the cosine function: \( \tan x = \frac{\sin x}{\cos x} \). Due to this definition, the tangent function is undefined whenever \( \cos x = 0 \), as division by zero is not possible. In such cases, the function has vertical asymptotes. These occur at odd multiples of \( \frac{\pi}{2} \) within the interval

• \( x = \frac{\pi}{2} \)
• \( x = \frac{3\pi}{2} \)

The tangent function has a period of \( \pi \), meaning that the pattern of the function repeats every \( \pi \) units. This is an important property when solving trigonometric equations, as it allows us to find all solutions within a given interval by considering just one period. For example, when solving \( \tan x = -1 \), solutions can be found at particular angles within each period of \( \pi \). In this problem, solutions are found at \( \frac{3\pi}{4} \) and \( \frac{7\pi}{4} \) within the interval \([0, 2\pi)\).
Understanding these properties is crucial when working with equations involving the tangent function, as they help identify all possible solutions.

Sine Function

The sine function, denoted as "\( \sin \)," is another core trigonometric function, which measures the vertical component of a point on the unit circle. The function takes angles (often in radians) as inputs and outputs values ranging from -1 to 1. The sine function has a distinct periodic nature with a period of \( 2\pi \). This means it returns to its starting value every \( 2\pi \) units.

• The sine function reaches its maximum value of 1 at \( x = \frac{\pi}{2} + 2n\pi \), where \( n \) is an integer.
• Conversely, it reaches its minimum value of -1 at \( x = \frac{3\pi}{2} + 2n\pi \).

For the problem in question, we solve the equation \( \sin x = 1 \), which gives a solution of \( x = \frac{\pi}{2} \) within the interval \([0, 2\pi)\). Since \( \sin x \) can only reach this value at one specific point within the cycle, it results in a single solution. Recognizing these key characteristics of the sine function is essential when solving trigonometric equations, as it aids in efficiently identifying appropriate solutions.

Solving Equations

When solving trigonometric equations, especially involving products like \((\tan x + 1)(\sin x - 1) = 0\), it is vital to understand how each factor contributes to finding solutions. Each part of the equation must be individually set equal to zero to find possible solutions, utilizing the zero-product property.
In simpler terms, if two numbers multiply to zero, then at least one of the numbers must be zero. This results in dealing with separate simpler equations:

• \( \tan x + 1 = 0 \) simplifies to \( \tan x = -1 \).
• \( \sin x - 1 = 0 \) simplifies to \( \sin x = 1 \).

Each of these simplified equations provides specific solutions within the defined interval \([0, 2\pi)\). After obtaining solutions from each equation:

• Combine them to form the complete set of solutions for the original equation.

For this particular problem, you deduce that the solutions are \( x = \frac{3\pi}{4}, \frac{7\pi}{4}, \) and \( \frac{\pi}{2} \). Using the zero-product property and understanding each trigonometric function's behavior are key strategies in trigonometric problem-solving.