Problem 35 - Algebra and Trigonometry with Analytic Geometry

Question

Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$ \sin 2 t+\sin t=0 $$

Step-by-Step Solution

Answer

The solutions are \( t = 0, \pi, \frac{2\pi}{3}, \frac{4\pi}{3} \).

Step 1: Simplify the Equation

Start by using the double angle identity for sine. Recall that \( \sin 2t = 2 \sin t \cos t \). Substitute this into the equation:\[2 \sin t \cos t + \sin t = 0\]

Step 2: Factor the Expression

Notice that \( \sin t \) is a common factor in the equation. Factor out \( \sin t \) from both terms:\[\sin t (2 \cos t + 1) = 0\]

Step 3: Solve Each Factor Separately

This equation is satisfied when either factor is zero. Thus, we set each factor equal to zero and solve.1. \( \sin t = 0 \)2. \( 2 \cos t + 1 = 0 \)

Step 4: Solve \( \sin t = 0 \)

For \( \sin t = 0 \), find the values of \( t \) in the interval \([0, 2\pi)\) where the sine function is zero. These occur at:\[ t = 0, \pi \]

Step 5: Solve \( 2 \cos t + 1 = 0 \)

First, solve for \( \cos t \) by isolating it:\[2 \cos t + 1 = 0 \2 \cos t = -1 \\cos t = -\frac{1}{2}\]The values of \( t \) where \( \cos t = -\frac{1}{2} \) in the interval \([0, 2\pi)\) are:\[ t = \frac{2\pi}{3}, \frac{4\pi}{3} \]

Step 6: Collect All Solutions

Combine the solutions from both factors. The solutions to the equation \( \sin 2t + \sin t = 0 \) in the interval \([0, 2\pi)\) are:- From \( \sin t = 0 \): \( t = 0, \pi \)- From \( \cos t = -\frac{1}{2} \): \( t = \frac{2\pi}{3}, \frac{4\pi}{3} \)Thus, the final solutions are: \( t = 0, \pi, \frac{2\pi}{3}, \frac{4\pi}{3} \).

Key Concepts

Double Angle Identity

The double angle identity for sine is a powerful tool in solving trigonometric equations, especially when dealing with expressions that involve double angles like \(2t\). This identity states that \( \sin 2t = 2 \sin t \cos t \). It's essentially a formula that helps us rewrite the function in a different, more workable form.

In our problem, the given equation \( \sin 2t + \sin t = 0 \) is simplified using the double angle identity. It allows us to transform the left side of the equation to \( 2 \sin t \cos t + \sin t = 0 \).

Using this identity aids in making the equation more approachable since it reveals a common factor that can be easily factored out later. Remembering and effectively utilizing this identity can significantly simplify the process of solving more complex trigonometric equations.

Sine Function

The sine function is an essential component of trigonometry. It describes a wave-like pattern and is periodic with a period of \(2\pi\), meaning it repeats every \(2\pi\) radians.

For the equation \( \sin t = 0 \), we look for values of \(t\) where the sine function crosses the horizontal axis. The points of zero are found at integer multiples of \(\pi\); however, within the interval \([0, 2\pi)\), the specific values are \(t = 0\) and \(t = \pi\).

Recognizing these key values where the sine function is zero is crucial, as they signify potential solutions to our trigonometric equations.

• Zero values: Occur at \(0, \pi, 2\pi, 3\pi, \ldots\).
• Sine's range: From -1 to 1.

Understanding the behavior of the sine function helps in predicting and calculating problem solutions effectively.

Factoring Trigonometric Expressions

Factoring is a critical technique when simplifying and solving trigonometric equations. It involves breaking down complex expressions into the product of simpler ones, which can then be solved separately.

In the example \( \sin t (2 \cos t + 1) = 0 \), this equation is structured such that we can apply the zero product property. This property states that if a product is zero, at least one of the factors must also be zero.

We start by identifying \( \sin t\) as a common factor in the equation \( 2 \sin t \cos t + \sin t = 0 \). Factoring \( \sin t\) results in \( \sin t (2 \cos t + 1) = 0 \).

The beauty of factoring here is that it splits the original problem into two simpler problems: \( \sin t = 0 \) and \( 2 \cos t + 1 = 0 \). Each can be solved independently, and their combined solutions offer the complete solution to the original equation.

• Look for common factors.
• Apply zero product property.

Mastering factoring allows you to tackle complex equations by breaking them down into manageable parts.