Chapter 7: Q. 59 (page 465)

In Problems 57– 80, solve each equation on the interval $0 \leq θ \leq 2 π$
role="math" localid="1646843196182" $2 \sin^{2} θ - \sin θ - 1 = 0$

Short Answer

Expert verified

The solution set of the given equation is $\{\frac{7 π}{6}, \frac{11 π}{6}, \frac{π}{2}\}$

Step by step solution

Step 1. Given Information

In the given problems we have to solve each equation on the interval $0 \leq θ \leq 2 π$

$2 \sin^{2} θ - \sin θ - 1 = 0$

Step 2. This equation is a quadratic equation (in sinθ) that can be factored.

$2 \sin^{2} θ - \sin θ - 1 = 0 2 \sin^{2} θ - (2 - 1) \sin θ - 1 = 0 2 \sin^{2} θ - 2 \sin θ + \sin θ - 1 = 0 2 \sin θ (\sin θ - 1) + 1 (\sin θ - 1) = 0 (2 \sin θ + 1) (\sin θ - 1) = 0$

Step 3. Use the Zero-Product Property.

$2 \sin θ + 1 = 0 or \sin θ - 1 = 0 2 \sin θ + 1 - 1 = 0 - 1 or \sin θ - 1 + 1 = 0 + 1 2 \sin θ = - 1 or \sin θ = 1 \frac{2}{2} \sin θ = - \frac{1}{2} or \sin θ = 1 \sin θ = - \frac{1}{2} or \sin θ = 1$

Step 4. Solving each equation in the interval [0,2π), we obtain

$θ = \frac{7 π}{6}, \frac{11 π}{6} or θ = \frac{π}{2}$

So the solution set is $\{\frac{7 π}{6}, \frac{11 π}{6}, \frac{π}{2}\}$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

In Problems 57– 80, solve each equation on the interval $0 \leq θ \leq 2 π$
role="math" localid="1646843196182" $2 \sin^{2} θ - \sin θ - 1 = 0$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. This equation is a quadratic equation (in sinθ) that can be factored.

Step 3. Use the Zero-Product Property.

Step 4. Solving each equation in the interval [0,2π), we obtain

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Theoretical and Mathematical Physics

Mechanics Maths

Calculus

Decision Maths

Applied Mathematics

Study anywhere. Anytime. Across all devices.

In Problems 57– 80, solve each equation on the interval 0≤θ≤2πrole="math" localid="1646843196182" 2sin2θ-sinθ-1=0

Short Answer

Step by step solution

Step 1. Given Information

Step 2. This equation is a quadratic equation (in sinθ) that can be factored.

Step 3. Use the Zero-Product Property.

Step 4. Solving each equation in the interval [0,2π), we obtain

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Theoretical and Mathematical Physics

Mechanics Maths

Calculus

Decision Maths

Applied Mathematics

Study anywhere. Anytime. Across all devices.

In Problems 57– 80, solve each equation on the interval $0 \leq θ \leq 2 π$
role="math" localid="1646843196182" $2 \sin^{2} θ - \sin θ - 1 = 0$