Problem 54 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

The solutions for \(x\) on the interval \([0,2 \pi]\) are \(x= \frac{3\pi}{4}, \frac{7\pi}{4}, \frac{\pi}{2}\)

Step 1: Set Each Factor Equal to Zero

Setting each of the two factors equal to zero, one finds two equations: \(\tan x+1=0\) and \(\sin x-1=0\) .

Step 2: Solve the Equation \(\tan x+1=0\)

To find the solution to the first equation, subtract 1 from both sides of the equation to get \(\tan x = -1\). The solutions to this equation in the interval \([0,2 \pi)\) are \(x = \frac{3\pi}{4}\) and \(x = \frac{7\pi}{4}\).

Step 3: Solve the Equation \(\sin x-1=0\)

To find the solution to the second equation, add 1 to both sides of the equation to get \(\sin x = 1\). The solution to this equation in the interval \([0,2 \pi)\) is \(x = \frac{\pi}{2}\).

Step 4: Combine Solutions

The solutions therefore from both these equations must be collated into a single solution set.

Key Concepts

Tangent Function

The tangent function, often denoted as \(\tan x\), is part of the primary trigonometric functions and is understood through the relationship \(\tan x = \frac{\sin x}{\cos x}\). This function produces a ratio derived from the opposite side to the adjacent side in a right triangle. However, it behaves uniquely in that it is periodic with a period of \(\pi\), meaning its behavior repeats every \(\pi\) radians.
Unlike sine and cosine, tangent can take any real number as a value from \(-\infty\) to \(+\infty\), resulting in vertical asymptotes every \(\frac{\pi}{2} + k\pi\) (where \(k\) is an integer), where it is undefined.

• When solving equations involving the tangent function, focus on reducing or simplifying the equation to one of its known values, such as \(\tan x = 0\), \(\tan x = 1\), or in our case, \(\tan x = -1\).
• A special property at \(\tan x = -1\) occurs at specific points like \(x = \frac{3\pi}{4}\) and \(x = \frac{7\pi}{4}\).

This knowledge helps us find solutions within specified intervals, which are often bounded due to the periodic nature of the function.

Sine Function

The sine function, noted as \(\sin x\), is one of the fundamental trigonometric functions and represents a ratio associated with a right triangle's hypotenuse. It's defined over all real numbers, repeating its values every \(2\pi\), known as periodicity.
It varies between \(-1\) and \(1\), reaching inclusive maximum and minimum values over its period. The function achieves these extremities at critical angles, where \(\sin x = 1\) at \(x = \frac{\pi}{2} + 2k\pi\) (where \(k\) is an integer).

• In our equation, setting \(\sin x = 1\) primarily gives us one solution in the interval \([0, 2\pi)\), specifically \(x = \frac{\pi}{2}\).
• Understanding the sine wave and its intersection with these values assists in predicting its behavior.

Recognizing these critical points allows the resolution of the sine component in trigonometric equations effectively.

Solution Intervals

When solving trigonometric equations, defining the solution interval is crucial to pinpoint the exact solutions. An interval like \([0, 2\pi)\) dictates that the solutions must be found within this range, inclusive of the first number but excluding the second.
This interval represents one full rotation angle measure in radians, where various key trigonometric function properties repeat.

• Given the periodicity, solutions for \(\tan x = -1\) and \(\sin x = 1\) are limited to their first occurrences within \([0, 2\pi)\).
• It’s vital to examine the interval to ensure no solutions fall out of bounds or are duplicated by periodic repeats.

Hence, understanding the impact of solving equations within specific intervals simplifies the solution merging process, ensuring completeness and accuracy.