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Precalculus(NEED HELP ASAP PLEASE!!)

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Solve:
sin(theta)+cos(theta) = 1
0 < or equal to theta < 2pi

Solution Set:__________ ?

  • Precalculus(NEED HELP ASAP PLEASE!!) -

    sin T +sqrt (1-sin^2T) = 1
    sqrt(1-sin^2T) = 1 - sin T
    1 - sin^2 T = 1 - 2 sin t + sin^2 T

    -sin^2 T = -2 sin T + sin^2 T
    2 sin^2 T = 2 sin T
    sin T = 1
    T = 0
    That tells you something. The only places this happens are when sin = 0 or cos = 0
    That is at 0, pi/2, pi, and 3 pi/2

  • Precalculus(NEED HELP ASAP PLEASE!!) -

    Oh, and then only when the one that is not zero is +1, not -1

  • Precalculus(NEED HELP ASAP PLEASE!!) -

    Damon I don't understand you 2nd post but is the answer: 0, pi/2, pi, and 3 pi/2 ??
    If it is.. then unfortunately its not the correct answer. Do you think you went wrong somewhere?

  • Precalculus(NEED HELP ASAP PLEASE!!) -

    sin T = +1 when T = 90 degrees , pi/2
    cos T = 1 when T = 0 or 2pi

    if T = 0 for example
    sin T = 0
    cos T = 1
    and sin T + cos T = 1
    So that just plain checks.

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