Precalculus with Limits: A Graphing Approach
Precalculus with Limits: A Graphing Approach
7th Edition
ISBN: 9781305071711
Author: Ron Larson
Publisher: Brooks/Cole
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Concept explainers

Ratios
A ratio is a comparison between two numbers of the same kind. It represents how many times one number contains another. It also represents how small or large one number is compared to the other.
Trigonometric Ratios
Trigonometric ratios give values of trigonometric functions. It always deals with triangles that have one angle measuring 90 degrees. These triangles are right-angled. We take the ratio of sides of these triangles.
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Chapter 5.5, Problem 13E
To determine

To calculate: The solution of the equation 4sinxcosx=1 in the interval [0,2π) .

Expert Solution & Answer
Check Mark

Answer to Problem 13E

The solution set of the equation 4sinxcosx=1 in the interval [0,2π) is {0,π12,5π12,13π12,17π12} .

Explanation of Solution

Given information:

The given equation is 4sinxcosx=1 .

Formula used:

Use the double angle formula,

  sin2x=2sinxcosx

Calculation:

Substitute the value of sin2x into given equation.

  4sinxcosx=2sin2x

Further simplify.

  2sin2x=1

The solutions in the interval [0,2π) , substitute the factor sin2x=12 .

  x=0,π12,5π12,13π12,17π12

The solution set is

  {0,π12,5π12,13π12,17π12} .

Verify the solutions using graphical utility:

Draw the graph for the equation 4sinxcosx=1 as shown in Figure(1).

  Precalculus with Limits: A Graphing Approach, Chapter 5.5, Problem 13E

Figure(1)

From the above, it is clear that {0,π12,5π12,13π12,17π12} are the solutions.

Therefore, the solution of the equation 4sinxcosx=1 in the interval [0,2π){0,π12,5π12,13π12,17π12} .

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Chapter 5.5, Problem 12E
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Chapter 5.5, Problem 14E

Chapter 5 Solutions

Precalculus with Limits: A Graphing Approach

Ch. 5.1 - Prob. 1ECh. 5.1 - Prob. 2ECh. 5.1 - Prob. 3ECh. 5.1 - Prob. 4ECh. 5.1 - Prob. 5ECh. 5.1 - Prob. 6ECh. 5.1 - Prob. 7ECh. 5.1 - Prob. 8ECh. 5.1 - Prob. 9ECh. 5.1 - Prob. 10E
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Trigonometric Ratios; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=9-eHMMpQC2k;License: Standard YouTube License, CC-BY