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2 questions .

  1. Aug 6, 2004 #1

    abc

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    2 questions .....

    1- solve :

    5 > x^2 >= -9
    : x belongs to R
    ----------------------------------------------
    2- solve :

    cos (x+30) - sin (2x) = 0
    45 >= x >= 0

    ----------------------------------------------
    thanx
    regards
    abc
     
  2. jcsd
  3. Aug 6, 2004 #2
    cos(x+30) = sin(2x) = cos (90-2x)

    this means : x+30 = 90-2x and x+30 = -90+2x

    Then solve to x and you are done...
    marlon
     
  4. Aug 6, 2004 #3
    x^2 >= -9 for a real number : this is not possible. You only have 5>x^2>=0. The solution : x < sqrt(5 ) and x>-sqrt(5)
     
  5. Aug 6, 2004 #4
    Huh? It's true for any real number x.
     
  6. Aug 7, 2004 #5

    hallo Huh?

    I meant that x^2 is always bigger or equal to zero, when x is considered to be a real number. This does not apply for complex numbers though...

    regards
    marlon
     
  7. Aug 7, 2004 #6

    HallsofIvy

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    Staff Emeritus
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    marlon: you originally said "x^2 >= -9 for a real number : this is not possible." Muzza's point was that it certainly is possible.

    You then noted "I meant that x^2 is always bigger or equal to zero".

    Yes: x^2>= 0 which is itself larger than -9. x^2>= -9 for all real numbers x.

    If you had said "that's always true" rather than "that's impossible" you would have been right: the solutions to 5>x^2>= -9 is exactly the set of numbers whose square is less than 5.
     
  8. Aug 8, 2004 #7
    Thank you HallsofIvy ;)
     
  9. Aug 8, 2004 #8

    OK I STAND CORRECTED

    regards
    marlon
     
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