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Math Help - Worked solution provided, but why is z = 360 deg rejected?

  1. #1
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    Worked solution provided, but why is z = 360 deg rejected?

    Can any kind soul help me in the below?

    Find z between 0 and 360 degrees which satisfy the equation.
    1. sec^2 z = 4 sec z - 3

    Answer:

    sec^2 z = 4 sec z - 3
    sec^2 z - 4 sec z + 3 = 0
    (sec z - 3)( sec z - 1) = 0
    sec z = 3 or sec z = 1
    cos z = 1/3 or cos z = 1
    z = 70.5 deg, z = 0, 360 deg(rejected)
    cos is positive => so in 1st or 4th quadrant
    z = 70.5, 360-70.5
    = 70.5, 289.5

    My question: why is z = 360 deg rejected?
    why is z = 0 deg not rejected?
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  2. #2
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    Quote Originally Posted by ppppp77 View Post

    My question: why is z = 360 deg rejected?
    why is z = 0 deg not rejected?
    It is not clear why in this case, was your domain [0,360) or [0,360] ?
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  3. #3
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    Quote Originally Posted by ppppp77 View Post
    Can any kind soul help me in the below?

    Find z between 0 and 360 degrees which satisfy the equation.
    1. sec^2 z = 4 sec z - 3

    Answer:

    sec^2 z = 4 sec z - 3
    sec^2 z - 4 sec z + 3 = 0
    (sec z - 3)( sec z - 1) = 0
    sec z = 3 or sec z = 1
    cos z = 1/3 or cos z = 1
    z = 70.5 deg, z = 0, 360 deg(rejected)
    cos is positive => so in 1st or 4th quadrant
    z = 70.5, 360-70.5
    = 70.5, 289.5

    My question: why is z = 360 deg rejected?
    why is z = 0 deg not rejected?
    oh my, pickslides, you are really awesome!!
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