Problem 28 - Precalculus Mathematics for Calculus

Question

An equation is given. (a) Find all solutions of the equation. (b) Find the solutions in the interval \([0,2 \pi)\) $$\csc 3 \theta=5 \sin 3 \theta$$

Step-by-Step Solution

Answer

The solutions of the equation are \( \theta = \frac{1}{3} \arcsin \left( \frac{1}{\sqrt{5}} \right) + \frac{2k\pi}{3} \) and similar expressions within \([0, 2\pi)\).

Step 1: Understand the Identity

The equation given is \( \csc 3\theta = 5 \sin 3\theta \). Recall that \( \csc 3\theta \) is the reciprocal of \( \sin 3\theta \), so \( \csc 3\theta = \frac{1}{\sin 3\theta} \).

Step 2: Simplify the Equation

Substitute the identity \( \csc 3\theta = \frac{1}{\sin 3\theta} \) into the equation: \( \frac{1}{\sin 3\theta} = 5 \sin 3\theta \).

Step 3: Multiply to Clear the Denominator

Multiply both sides by \( \sin 3\theta \) to clear the denominator:\[ 1 = 5 \sin^2 3\theta \]

Step 4: Solve for \( \sin^2 3\theta \)

Rearrange the equation to solve for \( \sin^2 3\theta \):\[ \sin^2 3\theta = \frac{1}{5} \]

Step 5: Find \( \sin 3\theta \)

Take the square root of both sides to solve for \( \sin 3\theta \):\[ \sin 3\theta = \pm \frac{1}{\sqrt{5}} \]

Step 6: Find \( 3\theta \) for Each Case

For \( \sin 3\theta = \frac{1}{\sqrt{5}} \), \( 3\theta = \arcsin \left( \frac{1}{\sqrt{5}} \right) + 2k\pi \) and \( 3\theta = \pi - \arcsin \left( \frac{1}{\sqrt{5}} \right) + 2k\pi \). For \( \sin 3\theta = -\frac{1}{\sqrt{5}} \), \( 3\theta = -\arcsin \left( \frac{1}{\sqrt{5}} \right) + 2k\pi \) and \( 3\theta = \pi + \arcsin \left( \frac{1}{\sqrt{5}} \right) + 2k\pi \).

Step 7: Solve for \( \theta \)

Divide each expression by 3 to find \( \theta \):- \( \theta = \frac{1}{3} \arcsin \left( \frac{1}{\sqrt{5}} \right) + \frac{2k\pi}{3} \)- \( \theta = \frac{\pi}{3} - \frac{1}{3} \arcsin \left( \frac{1}{\sqrt{5}} \right) + \frac{2k\pi}{3} \)- \( \theta = -\frac{1}{3} \arcsin \left( \frac{1}{\sqrt{5}} \right) + \frac{2k\pi}{3} \)- \( \theta = \frac{\pi}{3} + \frac{1}{3} \arcsin \left( \frac{1}{\sqrt{5}} \right) + \frac{2k\pi}{3} \)

Step 8: Find Solutions in [0, 2π)

Substitute different integer values for \( k \) to find all solutions for \( \theta \) within the interval \([0, 2\pi)\).

Key Concepts

Cosecant Function

The cosecant function, denoted as \( \csc \theta \), is one of the six fundamental trigonometric functions. It is the reciprocal of the sine function. This means that \( \csc \theta = \frac{1}{\sin \theta} \), making it undefined whenever the sine of the angle is zero.

• In this exercise, the equation \( \csc 3\theta = 5 \sin 3\theta \) uses the concept of the cosecant, highlighting its reciprocal relationship with sine.
• The conversion \( \csc 3\theta = \frac{1}{\sin 3\theta} \) simplifies solving the equation by eliminating the denominators.

Understanding this reciprocal nature is crucial when manipulating trigonometric identities, especially in solving equations where trigonometric functions are involved.

Arc Sine Function

The arc sine function, represented as \( \arcsin \), is the inverse of the sine function. It is used to determine the angle whose sine value is known.

• When \( \sin \theta = x \), then \( \theta = \arcsin(x) \). This provides the principal value of the angle.
• In our problem, we encounter \( \sin 3\theta = \frac{1}{\sqrt{5}} \), leading us to use \( 3\theta = \arcsin \left( \frac{1}{\sqrt{5}} \right) \).
• The arc sine function is crucial for resolving the angles that satisfy the equation as it provides the key steps in finding all potential angle solutions.

The function is defined only for values \(-1 \leq x \leq 1\), corresponding to angles within \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), facilitating the determination of specific angle solutions within known constraints.

Trigonometric Identities

Trigonometric identities are equations that are true for all values within the domains of the involved trigonometric functions. They play a significant role in simplifying trigonometric equations.

• Basic identities, such as \( \csc \theta = \frac{1}{\sin \theta} \), illustrate the interconnected nature of trigonometric functions.
• In this exercise, the identity enables us to convert \( \csc 3\theta \) into terms of \( \sin 3\theta \).
• By expressing the equation as \( \frac{1}{\sin 3\theta} = 5 \sin 3\theta \) and clearing denominators, it is simplified into \( 1 = 5 \sin^2 3\theta \).

Using these identities allows us to reframe complex equations into more manageable forms, leading to easier solutions. Knowing and applying the correct trigonometric identities is essential for successfully solving trigonometric equations.

Interval Notation

Interval notation is a mathematical notation used to describe a range of values between two endpoints. It is crucial for defining the domain or possible solutions for trigonometric equations.

• The interval \([0, 2\pi)\) specifies all angle solutions that must fit within one complete revolution around the circle, starting from 0 and just below \(2\pi\).
• In this context, after solving \( \theta = \frac{1}{3} \arcsin \left( \frac{1}{\sqrt{5}} \right) + \frac{2k\pi}{3} \), we need to substitute integer values of \( k \) to find solutions within this interval.
• Only solutions that fall within \([0, 2\pi)\) are valid as per the problem's requirement.

This form of notation is both concise and clear, making it an ideal method for expressing a range of solutions, particularly in cases involving periodic functions like trigonometry.