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Problem Check

  1. Jul 8, 2009 #1

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    1. The problem statement, all variables and given/known data

    Tan[tex]^{2}[/tex]X - 1 = 0

    2. Relevant equations



    3. The attempt at a solution

    I added 1 to both sides, then I took the square root of both sides giving me:

    tan x = [tex]\pm[/tex]1

    which gives me answers at pi/4, 3pi/4, 5pi/4 and so on.

    Do I just write pi/4 + 2kpi as my answer?
     
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  3. Jul 8, 2009 #2

    Cyosis

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    All the answers you've listed are correct, however depending on what k is incomplete. I assume k is an integer. If so then your final answer are points that are all on the same 'spot' of the unit circle. There are more spots where the tangent is 1 or -1.
     
    Last edited: Jul 8, 2009
  4. Jul 8, 2009 #3

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    How do I list every spot where tan = 1 or -1? without actually "listing" them all out.
     
  5. Jul 8, 2009 #4

    Cyosis

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    What you're doing now is making 2Pi rotations, you could also make pi rotations or pi/2 rotations or pi/3 rotations etc. You need to find the angle that fits your equation.

    What is the difference between pi/4, 3pi/4, 5pi/4, etc?
     
  6. Jul 8, 2009 #5

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    OH, so for this specific problem I'm going to write:

    pi/4, 3pi/4, 5pi/4, 7pi/4, + pi/2? In that order?
     
  7. Jul 8, 2009 #6
    No, you can start with pi/4 + something with k where, for every integer k, you will get pi/4, 3pi/4, 5pi/4, 7pi/4, etc. as well as the negative solutions -pi/4, -3pi/4, -5pi/4, -7pi/4, etc.
     
    Last edited: Jul 8, 2009
  8. Jul 8, 2009 #7

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    SO my final answer will be:


    pi/4 + 2pi/4?


    How would I do this for a repeating angle such as:

    sin x = -1/2 or 1/2 with no limits? since the difference are different.

    Like pi/6, 5pi/6, 7pi/6.

    The differences are different.
     
  9. Jul 8, 2009 #8
    As Cyosis said, What is the difference between pi/4, 3pi/4, 5pi/4, etc?
    The solution will be pi/4 + k*(common difference between each of the numbers above)

    For a problem like sin2x = 1/4 or sin x = -1/2, 1/2 where the differences are different, you give "two" solutions
    x = pi/6 + 2k*pi and x = 5pi/6 + 2k*pi
    or sometimes as
    x = pi/6 + 2k*pi, 5pi/6 + 2k*pi

    Make sense?
     
    Last edited: Jul 8, 2009
  10. Jul 8, 2009 #9

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    Definitely ! Thanks.
     
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