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Math Help - Solving for X

  1. #1
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    Solving for X

    Given the curve y=1-2cos^2x+cosx for 0<x<2\pi. Calculate the values of x for which the curve meets the x-axis.

    I don't know where I went wrong but this is my attempt, textbook answer is x=\frac{4\pi}{3} and \frac{2\pi}{3}

    Attempt

    Meet x-axis,  y=0

    1-2cos^2x+cosx=0

    (2cosx+1)(-cosx+1)=0

    cosx=\frac{-1}{2}[Q_2,Q_3] , cosx=1[Q_1,Q_4]

    \alpha=2.0943rad & \alpha=0

     x=\pi-2.0943, x=\pi+2.0943, x=0(rej), x=2\pi(rej)

    = 1.05rad          = 5.23rad
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  2. #2
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    Quote Originally Posted by Punch View Post
    Given the curve y=1-2cos^2x+cosx for 0<x<2\pi. Calculate the values of x for which the curve meets the x-axis.

    I don't know where I went wrong but this is my attempt, textbook answer is x=\frac{4\pi}{3} and \frac{2\pi}{3}

    Attempt

    Meet x-axis,  y=0

    1-2cos^2x+cosx=0

    (2cosx+1)(-cosx+1)=0

    cosx=\frac{-1}{2}[Q_2,Q_3] , cosx=1[Q_1,Q_4] your work is fine to here

    \alpha=2.0943rad & \alpha=0 these are 2 of 4 solutions which include 0 and 360 degrees.

     x=\pi-2.0943, x=\pi+2.0943, x=0(rej), x=2\pi(rej)

    = 1.05rad          = 5.23rad
    Hi Punch,

    Cosx is the horizontal co-ordinate of a point on the circle of radius =1 if you like, centred at the origin.

    When you obtain arccos using the value -0.5,
    your calculator will return one of two angles this corresponds to.

    You wrote your angles values "without" \pi.

    It's much simpler to approach this type of question,
    by considering cosx to be a horizontal co-ordinate of value -0.5,
    to the left of the origin.

    This co-ordinate references 2 angles, one less than 180 degrees
    and the other greater than 180 degrees.

    If you calculate arccos(0.5) you obtain the acute angle that you add to or subtract from 180 degrees.

    Note that the question excludes x=0 or x=360 degrees,
    so it is asking you to find the 2 angles x, referenced by cosx=-0.5.
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  3. #3
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    so does that mean that my answer is right?
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  4. #4
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    thanks i understood it
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  5. #5
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    Hi Punch,

    the first value of \alpha that you calculated is one value of the angle x.
    You should not be combining that value with \pi
    Attached Thumbnails Attached Thumbnails Solving for X-cosx.jpg  
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  6. #6
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    I was working out the value of the basic angle.
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  7. #7
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    Hi Punch,

    to get the "acute" angle, evaluate arccos(0.5), ie cos^{-1}0.5

    rather than arccos(-0.5).

    Then you can add to and subtract from 180 degrees to obtain x.

    When you evaluate arccos(-0.5), calculators typically give one of the
    two angles x. You still need to calculate the other one.
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  8. #8
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    Thanks, I spotted that
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