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Trignometric 'equation'

  1. Oct 12, 2005 #1
    hey
    first of all, that s not an equation, i dont know the word in english (sorry)
    but here... i have to solve that in the interval [-2pi , pi ]

    cos(2x)-√3sin(2x) ≥ -√2

    here is what i did.....
    2(.5cos(2x)-√3 /2 sin(2x) ≥ -√2
    cos(pi/3)cos(2x)-sin(pi/3)sin(2x) ≥ -√2 /2
    cos(pi/3 + 2x) ≥ -√2 / 2
    and that s gonna be....
    pi/3 + 2x ≥ pi/4

    x ≥ -pi/24

    now what, is that all what i have to do? is waht i did right? and finally, what should the solution to this 'problem' be ????

    thanks!
     
  2. jcsd
  3. Oct 12, 2005 #2

    mezarashi

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    Homework Helper

    For the record, these are called inequalities.

    To your question, do you think it's the right answer?

    consider: cos z ≥ -√2 / 2
    where z = pi/3 + 2x

    For what values of z is this inequality satisfied? If you know the possible values of z, can you determine the possible values of x?
     
    Last edited: Oct 12, 2005
  4. Oct 12, 2005 #3
    sorry, but I didnt get what you mean... All what I got from what you said is that x should be bigger or equal than (-3√2 - 2pi)/12
    :huh:
     
  5. Oct 12, 2005 #4
    oh, the possible values for z are 5pi/6 , -5pi/6 , -7pi/6 i think that is it for the interval i have (-2pi , pi)
    so now what ?? :confused:
     
  6. Oct 12, 2005 #5
    hmmm, is that right:

    the possible values of x are:
    pi/6
    -3pi/4
    -7pi/12

    are those the solutions for this INEQUALITY ??
     
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