Problem 9 - 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. $$2 \csc x+4=\csc x+6$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).

Step 1: Isolate the Trigonometric Function

First, let's start by isolating the trigonometric function in the equation. We have: \[2 \csc x + 4 = \csc x + 6\]Subtract \( \csc x \) from both sides to get: \[csc x + 4 = 6\]

Step 2: Solve for the Cosecant

Next, isolate \( \csc x \) by subtracting 4 from both sides: \[\csc x = 2\]

Step 3: Convert Cosecant to Sine

Remember that \( \csc x = \frac{1}{\sin x} \). Knowing this, set up the equation: \[\frac{1}{\sin x} = 2\]Solve for \( \sin x \) by taking the reciprocal: \[\sin x = \frac{1}{2}\]

Step 4: Find Solutions in the Given Interval

We need to find \( x \) such that \( \sin x = \frac{1}{2} \) within the interval \([0, 2\pi)\). The sine function equals \( \frac{1}{2} \) at \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \). Therefore, the solutions are: \[x = \frac{\pi}{6}, \frac{5\pi}{6}\]

Key Concepts

Cosecant Function

The cosecant function, denoted as \( \csc x \), is one of the reciprocal trigonometric functions. It's closely related to the sine function, since \( \csc x \) is defined as \( \frac{1}{\sin x} \). This means that wherever the sine function is defined and not equal to zero, the cosecant function will also be defined.

• When \( \sin x = 1 \), \( \csc x = 1 \).
• When \( \sin x = \frac{1}{2} \), \( \csc x = 2 \).

It's important to note that when \( \sin x = 0 \), the value of \( \csc x \) becomes undefined, because division by zero is not possible. Thus, the cosecant is crucial for problems where solutions involve angles corresponding to specific values of sine.

Sine Function

The sine function, represented as \( \sin x \), is one of the fundamental trigonometric functions. It indicates the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. The sine function is periodic, repeating its values every \( 2\pi \).

• The values of \( \sin x \) range from -1 to 1.
• It reaches 1 at \( x = \frac{\pi}{2} + 2k\pi \), where \( k \) is an integer.
• It reaches -1 at \( x = \frac{3\pi}{2} + 2k\pi \).
• It equals 0 at \( x = k\pi \).

In solving trigonometric equations, converting expressions into terms of sine can simplify finding solutions within a given interval. Like in the exercise, finding when \( \sin x = \frac{1}{2} \) helps pinpoint the angle values that fulfill the trigonometric equation.

Interval Notation

Interval notation is a concise way of expressing a range of numbers, often used in mathematics to specify the domain where a function or equation is applicable. In the context of trigonometric equations, it tells us where we should search for valid solutions. In this exercise, the interval is given as \([0, 2\pi)\).

• The \([\) indicates that \( 0 \) is included in the interval.
• The \(()\) indicates that \( 2\pi \) is not included.

Using interval notation helps to systematically determine which angles fall within the allowable range for a trigonometric solution. It’s especially useful when dealing with periodic functions like sine, as it helps limit solutions to those within the specified cycle.