Problem 7 - Algebra and Trigonometry

Question

\(3-16 \cdot\) Solve the given equation. $$ 2 \sin 2 \theta-3 \sin \theta=0 $$

Step-by-Step Solution

Answer

The solutions are \( \theta = n\pi \), where \( n \) is an integer.

Step 1: Factor out common terms

Notice that both terms in the equation contain a sine function. Let's factor out \( \sin \theta \) from the equation: \[ \sin \theta (2 \sin \theta - 3) = 0 \]

Step 2: Solve the factored equation

The factored equation presents us with two separate equations to solve: 1. \( \sin \theta = 0 \) 2. \( 2 \sin \theta - 3 = 0 \) Next, we'll solve each one individually.

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

The sine of an angle is zero at integer multiples of \( \pi \), i.e., \[ \theta = n\pi \] where \( n \) is an integer.

Step 4: Solve \( 2 \sin \theta - 3 = 0 \)

Rearrange the equation to solve for \( \sin \theta \): \[ 2 \sin \theta = 3 \] \[ \sin \theta = \frac{3}{2} \]Since the sine function has a range of \([-1, 1]\), there are no solutions to \( \sin \theta = \frac{3}{2} \).

Step 5: Compile the Solution

The only solutions come from Step 3, giving us the general solution. Thus, \( \theta = n \pi \), where \( n \) is any integer.

Key Concepts

Sine function

The sine function, denoted as \( \sin \theta \), is a fundamental concept in trigonometry. It relates to the angle of a right triangle to the ratio of the opposite side over the hypotenuse. The sine function is periodic and has a set range of values. - Range: The sine of an angle always falls between -1 and 1, inclusive. This means it's impossible for \( \sin \theta = \frac{3}{2} \), as that value is outside the range.- Periodicity: The sine function repeats every \( 2\pi \). This means if \( \sin \theta = a \), then \( \sin(\theta + 2k\pi) = a \) for any integer \( k \).Understanding these aspects of the sine function helps in solving trigonometric equations by recognizing valid solutions and eliminating impossibilities due to range constraints.

Factoring equations

Factoring is a useful technique in solving algebraic equations, including those involving trigonometric functions. The process involves rewriting an equation as a product of simpler expressions, making it easier to solve.In the context of the equation \( 2 \sin 2 \theta - 3 \sin \theta = 0 \), factoring involves identifying common terms. Here, \( \sin \theta \) is a common factor:- **Identify common terms:** In our equation, both terms on the left side involve \( \sin \theta \). - **Factor out the common term:** This turns the equation into \( \sin \theta (2 \sin \theta - 3) = 0 \).Factoring transforms a difficult problem into simpler parts, allowing us to solve each part separately. In this case, it provides two equations: \( \sin \theta = 0 \) and \( 2 \sin \theta - 3 = 0 \).

General solution

The general solution of a trigonometric equation provides a way to describe all possible solutions for a particular angle \( \theta \). Since trigonometric functions are periodic, they often have infinitely many solutions that can be generalized with a pattern.- **Solving \( \sin \theta = 0 \):** The angle \( \theta \) at which the sine function is zero occurs at integer multiples of \( \pi \). So, the general solution is \( \theta = n\pi \), where \( n \) is any integer.- **Implications:** This implies that the sine wave crosses the x-axis at these regular intervals.Given the periodic nature of sine, expressing solutions generally covers all possible instances of \( \theta \) that satisfy the equation. Understanding the periodic properties of the sine function helps in arriving at this general solution efficiently. In our specific problem, since \( \sin \theta = \frac{3}{2} \) has no real solutions, our general solution is solely \( \theta = n \pi \).