Trigonometric Equations...3

Discussion in 'Geometry and Trigonometry' started by nycmathguy, Dec 28, 2021.

  1. nycmathguy

    nycmathguy

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    Section 5.4

    Screenshot_20211226-094501_Samsung Notes.jpg

    Question 74

    sin(x + pi/2) - cos^2 x = 0

    Let sin(x + pi/2) = cos x

    cos x - cos^2 x = 0

    Factor

    cos x(1 - cos x ) = 0

    Set each factor to 0 and solve for x.

    cos x = 0

    x = pi/2, 3pi/2

    Set the other factor to 0 and solve for x.

    1 - cos x = 0

    -cos x = -1

    cos x = -1/-1

    cos x = 1

    x = 2pi

    You say?
     
    nycmathguy, Dec 28, 2021
    #1
  2. nycmathguy

    Country Boy

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    I would not say "Let sin(x + pi/2) = cos x". That is an identity, you do not need to "let" it be true in this particular case.
     
    Country Boy, Dec 28, 2021
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  3. nycmathguy

    nycmathguy

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    Is my answer correct?
     
    nycmathguy, Dec 28, 2021
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  4. nycmathguy

    Country Boy

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    cos(0)= 1 so x= 0 is an solution. If you want all solutions, since sine and cosine are periodic with period 2pi, a multiples of 2p:
    x= pi/2+ 2npi
    x= 3pi/2+ 2npi
    x= 2npi
    for any integer, n.
     
    Country Boy, Dec 28, 2021
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    nycmathguy likes this.
  5. nycmathguy

    nycmathguy

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    I was right, then.
     
    nycmathguy, Dec 28, 2021
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