Problem 38 - Precalculus Mathematics for Calculus

Question

Find all solutions of the given equation. $$\csc ^{2} \theta-4=0$$

Step-by-Step Solution

Answer

Solutions: \( \theta = \frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi, \frac{7\pi}{6} + 2k\pi, \frac{11\pi}{6} + 2k\pi \); \( k \in \mathbb{Z} \).

Step 1: Understand the Problem

We are given the equation \( \csc^2 \theta - 4 = 0 \). The goal is to find all the values of \( \theta \) (angles) that satisfy this equation.

Step 2: Rearrange the Equation

We start by rearranging the equation \( \csc^2 \theta - 4 = 0 \) to \( \csc^2 \theta = 4 \). This can be achieved by adding 4 to both sides of the equation.

Step 3: Solve for Cosecant

The relation \( \csc^2 \theta = 4 \) implies that \( \csc \theta = \pm 2 \). Recall that \( \csc \theta = \frac{1}{\sin \theta} \), so we rewrite the equation as \( \frac{1}{\sin \theta} = \pm 2 \).

Step 4: Solve for Sine

We rearrange \( \frac{1}{\sin \theta} = 2 \) to obtain \( \sin \theta = \frac{1}{2} \) and \( \frac{1}{\sin \theta} = -2 \) to obtain \( \sin \theta = -\frac{1}{2} \).

Step 5: Find Angles Satisfying \(\sin \theta = \frac{1}{2}\)

The angles \( \theta \) that satisfy \( \sin \theta = \frac{1}{2} \) are \( \theta = \frac{\pi}{6} + 2k\pi \) and \( \theta = \frac{5\pi}{6} + 2k\pi \), where \( k \) is an integer.

Step 6: Find Angles Satisfying \(\sin \theta = -\frac{1}{2}\)

The angles \( \theta \) that satisfy \( \sin \theta = -\frac{1}{2} \) are \( \theta = \frac{7\pi}{6} + 2k\pi \) and \( \theta = \frac{11\pi}{6} + 2k\pi \), where \( k \) is an integer.

Key Concepts

Cosecant Function

The cosecant function, denoted as \( \csc \theta \), is a fundamental part of trigonometry. It is related to the sine function and can be defined as the reciprocal of the sine of an angle:

• \( \csc \theta = \frac{1}{\sin \theta} \)

This means the cosecant function is undefined when the sine of the angle is zero, as division by zero is not possible.

Cosecant is particularly useful in trigonometric equations like the one in our problem \( \csc^2 \theta - 4 = 0 \). By understanding that \( \csc \theta \) is simply the "inverse" of \( \sin \theta \), we can reconfigure trigonometric identities to solve these types of equations.

By squaring the function \( \csc \theta \), we handle the equation using known values of \( \csc \theta \), and eventually relate them back to solve for known sine values.

Sine Function

The sine function, denoted as \( \sin \theta \), is a critical function in trigonometry defining the ratio of the length of the side opposite to an angle in a right triangle to the length of the triangle’s hypotenuse:

• \( \sin \theta = \text{opposite side} / \text{hypotenuse} \)

It is a periodic function, meaning it repeats its values in regular intervals, and is commonly used in solving for angles because of its predictable cyclical nature.

In our exercise, the sine function comes into play when we solve \( \csc \theta = 2 \) and \( \csc \theta = -2 \). Rearranging these gives us \( \sin \theta = \frac{1}{2} \) and \( \sin \theta = -\frac{1}{2} \). These are familiar sine values that correspond to specific angles.

Knowing these base angles helps in determining all possible solutions, leveraging the sine function's periodicity in the unit circle approach, which is critical in fully solving the given equation.

Angle Solutions

Finding all angle solutions for trigonometric equations involves understanding the unit circle, where angles are measured in radians. Sine's periodic nature means that angles repeating every \( 2\pi \) are potential solutions.

For \( \sin \theta = \frac{1}{2} \), the angles are known:

• \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \)

These angles reflect when sine reaches \( \frac{1}{2} \) in the first and second quadrants of the unit circle.
Adding \( 2k\pi \) where \( k \) is any integer, accounts for the periodic repeat of the sine wave.

Similarly, for \( \sin \theta = -\frac{1}{2} \), the angles are:

• \( \theta = \frac{7\pi}{6} \) and \( \theta = \frac{11\pi}{6} \)

These angles occur in the third and fourth quadrants, where the sine value is negative.
Including \( 2k\pi \) helps cover all repetitions of the wave, thus providing an exhaustive set of solutions for the original trigonometric equation.