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 17E
To determine

To calculate: The solution of the equation sin4x=−2sin2x in the interval [0,2π) .

Expert Solution & Answer
Check Mark

Answer to Problem 17E

The solution set of the equation sin4x=−2sin2x in the interval [0,2π) is {π2,7π6,11π6} .

Explanation of Solution

Given information:

The given equation is sin4x=−2sin2xsin4x+2sin2x=0 .

Formula used:

Use the double angle formula.

  sin2x=2sinxcosx

Calculation:

Substitute the value of sin4x in given equation.

  sin4x+2sin2x=2sin2xcos2x+2sin2x=2sin2x(cos2x+1)

Further simplify.

  2sin2x(cos2x+1)=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+1=0 .

  x=π2,3π2

Therefore, the solution set is

  {0,π2,π,3π2}∪{π2,3π2}={0,π2,π,3π2}

Verify the solutions using graphical utility:

Draw the graph for the equation sin4x+2sin2x=0 as shown in Figure(1).

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 5.5, Problem 17E

Figure(1)

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

Therefore, the solution of the equation sin4x=−2sin2x in the interval [0,2π){0,π2,π,3π2} .

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Chapter 5.5, Problem 16E
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Chapter 5.5, Problem 18E

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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