PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Chapter 5.5, Problem 18E
To determine

To calculate:Thesolution of the equation (sin2x+cos2x)2=1 in the interval [0,2π) .

Expert Solution & Answer
Check Mark

Answer to Problem 18E

The solution set of the equation (sin2x+cos2x)2=1 in the interval [0,2π) is {0,π4,π2,3π4,π,5π4,3π2,7π4} .

Explanation of Solution

Given information:

The given equation is (sin2x+cos2x)2=1 .

Formula used:

Use the formula.

  (a+b)2=a2+b2+2ab

Calculation:

Substitute the value sin2x for a and cos2x for b in the equation.

  sin22x+cos22x+2sin2xcos2x=11+2sin2xcos2x=12sin2xcos2x=0

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

  x=0,π2,π,3π2

The solutions in the interval [0,2π) , substitute the factor cos2x=0 .

  x=π4,3π4,5π4,7π4

Therefore, the solution set is

  {0,π2,π,3π2}∪{π4,3π4,5π4,7π4}={0,π4,π2,3π4,π,5π4,3π2,7π4}

Verify the solutions using graphical utility:

Draw the graph for the equation (sin2x+cos2x)2=1 as shown in Figure(1).

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 5.5, Problem 18E

Figure(1)

From the above, it is clear that {0,π4,π2,3π4,π,5π4,3π2,7π4} are the solutions.

Therefore, the solution of the equation (sin2x+cos2x)2=1 in the interval [0,2π){0,π4,π2,3π4,π,5π4,3π2,7π4} .

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Chapter 5.5, Problem 17E
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Chapter 5.5, Problem 19E

Chapter 5 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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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Find the solutions to a trig equation between 0 and 2pi; Author: Brian McLogan;https://www.youtube.com/watch?v=h7trDHjKCYc;License: Standard YouTube License, CC-BY