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 14E
To determine

To calculate:Thesolution of the equation sin2xsinx=cosx in the interval [0,2π) .

Expert Solution & Answer
Check Mark

Answer to Problem 14E

The solution set of the equation sin2xsinx=cosx in the interval [0,2π) is {0,π4,π2,3π4,5π4,3π2,7π4} .

Explanation of Solution

Given information:

The given equation is sin2xsinx=cosx .

Formula used:

Use the double angle formula.

  sin2x=2sinxcosx

Calculation:

Substitute the value of sin2x in the given equation.

  sin2xsinx=2sinxcosxsinx=2sin2xcosx

Further simplify.

  2sin2xcosx=cosx2sin2x=1sin2x=12

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

  x=0,π4,π2,3π4,5π4,3π2,7π4

The solution set is

  {0,π4,π2,3π4,5π4,3π2,7π4} .

Verify the solutions using graphical utility:

Draw the graph for the equation sin2xsinx=cosx as shown in Figure(1).

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 5.5, Problem 14E

Figure(1)

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

Therefore, the solution of the equation sin2xsinx=cosx in the interval [0,2π){0,π4,π2,3π4,5π4,3π2,7π4} .

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Chapter 5.5, Problem 15E

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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