Q. 64 In Problems 57鈥� 80, solve each... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 64 (page 465)

In Problems 57鈥� 80, solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$ .
$\cos^{2} 胃 - \sin^{2} 胃 + \sin 胃 = 0$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information

In the given problems we have to solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$ .

$\cos^{2} 胃 - \sin^{2} 胃 + \sin 胃 = 0$

Step 2. The equation in its present form contains a cosine and sine.

However, a form of the Pythagorean Identity, $\sin^{2} 胃 + \cos^{2} 胃 = 1$ , can be used to transform the equation into an equivalent one containing only cosines.

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

Step 3. Now factor the equation.

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

Use the Zero-Product Property.

role="math" localid="1646929056145" $2 \sin 胃 + 1 = 0 or \sin 胃 - 1 = 0 2 \sin 胃 + 1 - 1 = 0 - 1 or \sin 胃 - 1 + 1 = 0 + 1 2 \sin 胃 = - 1 or \sin 胃 = 1 \frac{2}{2} \sin 胃 = - \frac{1}{2} or \sin 胃 = 1 \sin 胃 = - \frac{1}{2} or \sin 胃 = 1$

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

$胃 = \frac{7 蟺}{6}, \frac{11 蟺}{6} or 胃 = \frac{蟺}{2}$

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

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

In Problems 57鈥� 80, solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$ .
$\cos^{2} 胃 - \sin^{2} 胃 + \sin 胃 = 0$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. The equation in its present form contains a cosine and sine.

Step 3. Now factor the equation.

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

Applied Mathematics

Discrete Mathematics

Statistics

Calculus

Mechanics Maths

Study anywhere. Anytime. Across all devices.

91影视

In Problems 57鈥� 80, solve each equation on the interval 0鈮�胃鈮�2蟺.cos2胃-sin2胃+sin胃=0

Short Answer

Step by step solution

Step 1. Given Information

Step 2. The equation in its present form contains a cosine and sine.

Step 3. Now factor the equation.

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

Applied Mathematics

Discrete Mathematics

Statistics

Calculus

Mechanics Maths

Study anywhere. Anytime. Across all devices.

In Problems 57鈥� 80, solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$ .
$\cos^{2} 胃 - \sin^{2} 胃 + \sin 胃 = 0$