Problem 8 - Precalculus Mathematics for Calculus

Question

Solve the given equation. $$3 \sin 2 \theta-2 \sin \theta=0$$

Step-by-Step Solution

Answer

Solutions: \( \theta = n\pi \) and \( \theta = 2n\pi \pm \arccos (1/3) \).

Step 1: Introduce the Double Angle Formula

We start by using the double angle formula for sine. Rewrite \( \sin 2\theta \) as \( 2 \sin \theta \cos \theta \). The equation becomes: \[ 3(2 \sin \theta \cos \theta) - 2 \sin \theta = 0 \] which simplifies to \[ 6 \sin \theta \cos \theta - 2 \sin \theta = 0 \].

Step 2: Factor the Equation

Factor out \( \sin \theta \) from the equation: \[ \sin \theta (6 \cos \theta - 2) = 0 \].

Step 3: Solve for \( \sin \theta = 0 \)

Consider the factor \( \sin \theta = 0 \). The solutions for this are \( \theta = n\pi \), where \( n \) is any integer.

Step 4: Solve for \( 6 \cos \theta - 2 = 0 \)

Now solve the equation \( 6 \cos \theta - 2 = 0 \). Isolate \( \cos \theta \) to get \[ \cos \theta = \frac{1}{3} \].

Step 5: Find General Solutions for \( \cos \theta = \frac{1}{3} \)

The solutions for \( \cos \theta = \frac{1}{3} \) are given by \( \theta = 2n\pi \pm \arccos \left( \frac{1}{3} \right) \), where \( n \) is any integer.

Key Concepts

double angle formula

In trigonometry, the double angle formulae are a crucial tool for simplifying expressions or solving equations that involve trigonometric functions. The double angle formula for sine is:

• \( \sin 2\theta = 2 \sin \theta \cos \theta \)

This formula allows us to express the sine of a double angle in terms of products of sine and cosine of the angle itself.
In the given equation, \(3 \sin 2\theta - 2 \sin \theta = 0\), applying the double angle formula converts \(\sin 2\theta\) into \(2 \sin \theta \cos \theta\).
This operation transforms the equation into a form that can be handled more easily—by combining or factoring common terms. It's an essential step for simplifying our trigonometric equation, paving the path for potential solutions.
Understanding how to use double angle formulas efficiently can simplify complex trigonometric functions, and is often your first step in breaking down problems like this.

factoring equations

Factoring is a powerful technique commonly used in algebra and trigonometry to simplify equations and find solutions. After using the double angle formula on the given expression, we arrive at \(6 \sin \theta \cos \theta - 2 \sin \theta = 0\).
The key here is to recognize common factors. In this case, \( \sin \theta \) appears in both terms of the equation.
By factoring \( \sin \theta \) out of the expression, we get:

• \( \sin \theta (6 \cos \theta - 2) = 0 \)

This results in a product of terms equal to zero. For this equation to be true, one or both of those terms must be zero.
Factoring simplifies the equation into two separate cases to consider: \( \sin \theta = 0 \) and \( 6 \cos \theta - 2 = 0 \).
This approach breaks the problem into manageable pieces, making it easier to find the generalized solutions for \( \theta \).

general solutions

Finding general solutions for trigonometric equations includes determining all possible values of \( \theta \) that satisfy the equation. Let's review the two separate conditions obtained after factoring.
For \( \sin \theta = 0 \):

• The values are periodic at \( \theta = n\pi \), where \( n \) is an integer.

This solution reflects the cyclic nature of the sine function, appearing at every integer multiple of \( \pi \).
Now, for the other condition, \( 6 \cos \theta - 2 = 0 \):

• Solving gives \( \cos \theta = \frac{1}{3} \).

The general solution involves:

• \( \theta = 2n\pi \pm \arccos \left( \frac{1}{3} \right) \) with \( n \) as any integer.

This accounts for the periodic nature of cosine that repeats every \( 2\pi \), allowing us to catch every instance where \( \cos \theta = \frac{1}{3} \).
General solutions capture the entire set of answers in trigonometric equations, reflecting their periodic patterns and offering complete solutions.