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Aslan and Shelley are finding the solution for \(2 \sin ^{2} \theta=\sin \theta, 0 < \theta \leq \pi\) Here is their work. \(2 \sin ^{2} \theta=\sin \theta\) \(\frac{2 \sin ^{2} \theta}{\sin \theta}=\frac{\sin \theta}{\sin \theta} \quad\) Step 1 \(2 \sin \theta=1 \quad\) Step 2 \(\sin \theta=\frac{1}{2} \quad\) Step 3 \(\theta=\frac{\pi}{6}, \frac{5 \pi}{6} \quad\) Step 4 a) Identify the error that Aslan and Shelley made and explain why their solution is incorrect. b) Show a correct method to determine the solution for \(2 \sin ^{2} \theta=\sin \theta, 0 < \theta \leq \pi\)

Short Answer

Expert verified
Valid solutions are \( \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \pi \). Aslan and Shelley missed \( \theta = \pi \).

Step by step solution

01

Set the Equation to Zero

Rewrite the equation as follows:\[ 2 \sin^{2} \theta - \sin \theta = 0 \]
02

Factor the Equation

Factor the left side of the equation:\[ \sin \theta (2 \sin \theta - 1) = 0 \]
03

Set Each Factor to Zero

Set each factor equal to zero and solve for \( \theta \):1. \( \sin \theta = 0 \)2. \( 2 \sin \theta - 1 = 0 \)
04

Solve for \( \theta \) when \( \sin \theta = 0 \)

When \( \sin \theta = 0 \), \( \theta \) can be:\[ \theta = 0, \pi \]However, since the range given is \( 0 < \theta \leq \pi \), only \( \theta = \pi \) is valid.
05

Solve for \( \theta \) when \( 2 \sin \theta - 1 = 0 \)

Solve for \( \sin \theta \):\[ 2 \sin \theta - 1 = 0 \]\[ 2 \sin \theta = 1 \]\[ \sin \theta = \frac{1}{2} \]\( \theta \) can be:\[ \theta = \frac{\pi}{6}, \frac{5\pi}{6} \]
06

Combine All Valid Solutions

Combine all valid solutions within the given range \( 0 < \theta \leq \pi \):\[ \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \pi \]
07

Identify the Error

Aslan and Shelley's mistake is in dividing by \( \sin \theta \), which is undefined when \( \sin \theta = 0 \). They missed the solution \( \theta = \pi \) because they incorrectly handled the equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solving trigonometric equations
To solve equations involving trigonometric functions, the goal is to find the angles, typically denoted as \( \theta \), that satisfy the given conditions. These steps will guide you through the process of solving trigonometric equations:

Step 1: **Rewrite the Equation in Standard Form**
The first step is to manipulate the original equation into a standard form such as zero on one side. For example, for \( 2 \sin^{2} \theta = \sin \theta \), we rewrite it as:

\[ 2 \sin^{2} \theta - \sin \theta = 0 \]

Step 2: **Factor the Equation**
Factoring allows you to split the equation into simpler parts. For instance:

\[ \sin \theta (2 \sin \theta - 1) = 0 \]

Step 3: **Set Each Factor to Zero**
Once factored, set each factor equal to zero and solve for \( \theta \):

\[ \sin \theta = 0 \quad \text{and} \quad 2 \sin \theta - 1 = 0 \]

This approach simplifies the solving process by breaking down the equation into more manageable pieces.
factoring trigonometric expressions
Factoring trigonometric expressions follows similar principles to factoring algebraic expressions and is essential for simplifying equations. Here’s a step-by-step guide to factoring trigonometric equations:

**Step 1: Identify Common Factors**
Look for terms that can be factored out, just like you would in algebra. For \( 2 \sin^{2} \theta - \sin \theta = 0 \), both terms include \( \sin \theta \):

\[ \sin \theta (2 \sin \theta - 1) = 0 \]

**Step 2: Apply the Zero Product Property**
Set each factor equal to zero:

\[ \sin \theta = 0 \quad \text{and} \quad 2 \sin \theta - 1 = 0 \]

**Step 3: Solve Each Equation**
Solve for \( \theta \) in each resulting equation separately. This method efficiently isolates potential solutions. For example:

\[ 2 \sin \theta = 1 \quad \rightarrow \quad \sin \theta = \frac{1}{2} \]

Factoring simplifies solving by reducing complex expressions into simple, solvable components.
identifying valid solutions
Identifying valid solutions involves ensuring that solutions fit within the given constraints of the problem. For our equation, \(2 \sin ^{2} \theta=\theta, 0 < \theta \leq \pi\), certain steps are followed to ensure correctness:

**Step 1: Check Range Limitations**
Verify that the solved values of \( \theta \) fall within the specified range. For \( 0 < \theta \leq \pi \), values like \( \theta = 0 \) are excluded, while \( \theta = \pi \) is the edge case.

**Step 2: Validity of Solutions**
For \( \sin \theta = 0 \), within the specified range, the valid solution is \( \theta = \pi \).

**Step 3: Combine Valid Solutions**
List all angles that meet the conditions:

\[ \theta = \frac{\pi}{6}, \frac{5 \pi}{6}, \pi \]

Avoid missing solutions by scrutinizing how the problem constraints affect calculated values. Solving trigonometric equations accurately requires careful consideration of the defined range and any implied limitations from the initial manipulation of the equations.

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Most popular questions from this chapter

Without solving, determine the number of solutions for each trigonometric equation in the specified domain. Explain your reasoning. a) \(\sin \theta=\frac{\sqrt{3}}{2}, 0 \leq \theta < 2 \pi\) b) \(\cos \theta=\frac{1}{\sqrt{2}},-2 \pi \leq \theta < 2 \pi\) c) \(\tan \theta=-1,-360^{\circ} \leq \theta \leq 180^{\circ}\) d) \(\sec \theta=\frac{2 \sqrt{3}}{3},-180^{\circ} \leq \theta < 180^{\circ}\)

Yellowknife, Northwest Territories, and Crowsnest Pass, Alberta, lie along the \(114^{\circ} \mathrm{W}\) line of longitude. The latitude of Yellowknife is \(62.45^{\circ} \mathrm{N}\) and the latitude of Crowsnest Pass is \(49.63^{\circ} \mathrm{N}\). Consider Earth to be a sphere with radius \(6400 \mathrm{km}\). a) Sketch the information given above using a circle. Label the centre of Earth, its radius to the equator, and the locations of Yellowknife and Crowsnest Pass. b) Determine the distance between Yellowknife and Crowsnest Pass. Give your answer to the nearest hundredth of a kilometre. c) Choose a town or city either where you live or nearby. Determine the latitude and longitude of this location. Find another town or city with the same longitude. What is the distance between the two places?

Determine the exact values of the other five trigonometric ratios under the given conditions. a) \(\sin \theta=\frac{3}{5}, \frac{\pi}{2}<\theta<\pi\) b) \(\cos \theta=\frac{-2 \sqrt{2}}{3},-\pi \leq \theta \leq \frac{3 \pi}{2}\) c) \(\tan \theta=\frac{2}{3},-360^{\circ}<\theta<180^{\circ}\) d) \(\sec \theta=\frac{4 \sqrt{3}}{3},-180^{\circ} \leq \theta \leq 180^{\circ}\)

Determine the exact value of each expression. a) \(\cos 60^{\circ}+\sin 30^{\circ}\) b) \(\left(\sec 45^{\circ}\right)^{2}\) c) \(\left(\cos \frac{5 \pi}{3}\right)\left(\sec \frac{5 \pi}{3}\right)\) d) (tan \(\left.60^{\circ}\right)^{2}-\left(\sec 60^{\circ}\right)^{2}\) e) \(\left(\cos \frac{7 \pi}{4}\right)^{2}+\left(\sin \frac{7 \pi}{4}\right)^{2}\) f) \(\left(\cot \frac{5 \pi}{6}\right)^{2}\)

Convert each radian measure to degrees. Express your answers as exact values and as approximate measures, to the nearest thousandth. a) \(\frac{2 \pi}{7}\) b) \(\frac{7 \pi}{13}\) c) \(\frac{2}{3}\) d) 3.66 e) -6.14 f) -20

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