Problem 35 - Precalculus: Functions and Graphs

Question

Find all solutions of the equation. $$\sin 2 x(\csc 2 x-2)=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{n\pi}{2}, x = \frac{\pi}{12} + k\pi, x = \frac{5\pi}{12} + k\pi\).

Step 1: Identify the equation to solve

The equation we need to solve is \(\sin 2x (\csc 2x - 2) = 0\). This equation involves trigonometric functions and is factored into two parts: \(\sin 2x\) and \(\csc 2x - 2\).

Step 2: Solve the first part: \(\sin 2x = 0\)

To solve \(\sin 2x = 0\), we find the angles where the sine function is zero. The general solution for \(\sin \theta = 0\) is \(\theta = n\pi\), where \(n\) is an integer. Therefore, for \(2x\), we have: \[2x = n\pi\] which gives \[x = \frac{n\pi}{2}\].

Step 3: Solve the second part: \(\csc 2x - 2 = 0\)

First, recall that \(\csc 2x = \frac{1}{\sin 2x}\). So, setting \(\csc 2x = 2\), we have: \[\frac{1}{\sin 2x} = 2\] which implies \[\sin 2x = \frac{1}{2}\]. The angles where sine equals \(\frac{1}{2}\) are \(\theta = \frac{\pi}{6} + 2k\pi\) and \(\theta = \frac{5\pi}{6} + 2k\pi\). Therefore, we solve: \[2x = \frac{\pi}{6} + 2k\pi\] and \[2x = \frac{5\pi}{6} + 2k\pi\].

Step 4: Solve for \(x\) from \(\sin 2x = \frac{1}{2}\)

From \(2x = \frac{\pi}{6} + 2k\pi\), we have \[x = \frac{\pi}{12} + k\pi\]. From \(2x = \frac{5\pi}{6} + 2k\pi\), we have \[x = \frac{5\pi}{12} + k\pi\].

Step 5: Combine solutions

The general solutions to the equation \(\sin 2x (\csc 2x - 2) = 0\) are \(x = \frac{n\pi}{2}, x = \frac{\pi}{12} + k\pi\), and \(x = \frac{5\pi}{12} + k\pi\), where \(n\) and \(k\) are integers. Each series of solutions covers different sets of angles where either \(\sin 2x = 0\) or \(\csc 2x - 2 = 0\).

Key Concepts

Sine Function

The sine function, denoted as \( \sin \theta \), is one of the primary functions in trigonometry. It represents the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. The sine function is periodic with a period of \( 2\pi \), which means it repeats its values in cycles of \( 2\pi \). This periodic nature makes it useful for solving various equations, especially those that involve periodic phenomena.

• The sine function achieves a maximum value of 1 and a minimum of -1.
• The sine of an angle is zero at integer multiples of \( \pi \) (i.e., \( \pi, 2\pi, 3\pi, \ldots \)).

In the exercise, solving \( \sin 2x = 0 \) means we look for angles where the sine value is zero. Thus the solution is \( 2x = n\pi \), with \( n \) being any integer. Dividing by 2, we get \( x = \frac{n\pi}{2} \).
Understanding where sine reaches zero helps in finding specific solutions in trigonometric equations.

Cosecant Function

The cosecant function, \( \csc \theta \), is simply the reciprocal of the sine function. This means \( \csc \theta = \frac{1}{\sin \theta} \). Because it's a reciprocal, the cosecant function has undefined values whenever \( \sin \theta = 0 \), making it important to know where the sine function has these zeros.

• Cosecant is undefined at \( \theta = n\pi \), where \( n \) is an integer.
• \( \csc \theta \) reaches an extended range compared to sine, potentially tending toward positive or negative infinity when sine approaches zero.

In our problem, solving \( \csc 2x - 2 = 0 \) involves finding when the reciprocal of sine equals 2, which simplifies to \( \sin 2x = \frac{1}{2} \). This gives us critical points at angles where the sine function equals one-half, specifically at \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \).
Understanding how the cosecant function behaves at specific values is essential for dealing with these reciprocal relationships.

General Solution of Trigonometric Equations

Trigonometric equations often have solutions known as 'general solutions' due to the periodic nature of trigonometric functions. This means instead of singular solutions, we have an infinite set of solutions spaced by their period.
Much like how \( \sin x = 0 \) has solutions at multiples of \( \pi \), a general solution uses such properties to find all possible angles that satisfy the equation.
For the full equation \( \sin 2x (\csc 2x - 2) = 0 \), the general solution combines solutions from its parts:

• From \( \sin 2x = 0 \): \( x = \frac{n\pi}{2} \)
• From \( \csc 2x = 2 \): \( x = \frac{\pi}{12} + k\pi \) or \( x = \frac{5\pi}{12} + k\pi \)

In these expressions, \( n \) and \( k \) are integers and allow us to list all solutions comprehensively. Understanding this concept ensures one can express complex trigonometric solutions fully instead of just isolated results.