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Find union between the two of the solutions

  1. Apr 3, 2008 #1
    1. The problem statement, all variables and given/known data

    I solve the equation of one function, which comes out with two solutions:

    1. cosx=-1, x=(2k+1)[itex]\pi[/itex] ; 2. cosx=1, x=2k[itex]\pi[/itex] (k [itex]\in \mathbb{Z})[/itex]

    2. Relevant equations


    3. The attempt at a solution

    Now, we need to find union between the two of the solutions:

    {[itex]\pi + 2k\pi[/itex]}[itex]\cup[/itex]{[itex]2k\pi [/itex]}= ??

    What will be the solution of this one?
     
    Last edited: Apr 3, 2008
  2. jcsd
  3. Apr 3, 2008 #2

    Hootenanny

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    HINT: The first set is the set of all odd numbers [multiplied by [itex]\pi[/itex]] and the second is the set of all even numbers (including zero) [multiplied by [itex]\pi[/itex]].
     
    Last edited: Apr 3, 2008
  4. Apr 3, 2008 #3
    Ok, I understand. What's next? :D
     
  5. Apr 3, 2008 #4

    Hootenanny

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    Well the solution set is the set of all odd multiples of [itex]\pi[/itex] and all even multiples of [itex]\pi[/itex] (including zero), which is the set of all ....?

    If your still not sure, try writing out the first few allowed solutions.
     
  6. Apr 3, 2008 #5
    set of all numbers, which is k[itex]\pi[/itex]? Like this?

    What are those few allowed solutions?
     
    Last edited: Apr 3, 2008
  7. Apr 3, 2008 #6

    Hootenanny

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    The set of all solutions is [itex]\left\{k\pi\right\}\;\; ,\; k\in\mathbb{Z}[/itex], which is the set of all integers, not the set of all numbers.

    By a few allowed solutions are meant the first few numbers in each set.

    EDIT: You need to correct your itex delimiters to allow the thread to display properly.
     
    Last edited: Apr 3, 2008
  8. Apr 3, 2008 #7
    Can you give me some of that numbers?
    P.S I correct the tags.
     
  9. Apr 3, 2008 #8

    Hootenanny

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    The solutions are simply integer multiples of [itex]\pi[/itex] like I said previously,

    [tex]\left\{k\pi\right\} \; ,\; k\in\mathbb{Z} = \ldots , -3\pi, -2\pi, -\pi, 0 , \pi, 2\pi, 3\pi, \ldots[/tex]
     
  10. Apr 3, 2008 #9
    I understand. Thanks.
     
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