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Trignometric inequality problem

  1. Aug 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Given that -pi < x < pi, solve the following inequality in radians

    root(2) - 2sin(x-(pi/3)) < 0


    3. The attempt at a solution


    root(2) - 2sin(x-(pi/3)) < 0
    - 2sin(x-(pi/3)) < -root(2)
    sin(x-(pi/3)) > (root(2))




    -pi<x<-pi

    -pi - (pi/3) < x - pi/3 < pi - (pi/3)

    -4pi < x-(pi/3) < 2pi/3


    and I know that -5pi/4 and pi/4 fit the new range as a result of the translation pi/3 I just don't know what to do with them. Can someone please help?


    That is so far what I have got I do not know where to proceed from
     
  2. jcsd
  3. Aug 19, 2013 #2

    Mark44

    Staff: Mentor

    You have a mistake in the 3rd line. You can go from the first inequality directly to this inequality by adding 2sin(x - ##\pi/3##) to both sides.

    ##\sqrt{2} < 2sin(x - \pi/3)##
    ??? This is saying that ##-\pi## is less than itself, which is not true. How did you get this?
     
  4. Aug 19, 2013 #3
    oh sorry thats a typo its actually -pi < x < +pi and root(2)/2
     
  5. Aug 19, 2013 #4

    Mark44

    Staff: Mentor

    Can you solve sin(u) > ##\sqrt{2}/2##? There are lots of intervals that satisfy this inequality. Once you get that, then solve for x, where u = x - ##\pi/3##.
     
  6. Aug 19, 2013 #5
    Is there a way to solve this without the use of a graph though?
     
  7. Aug 19, 2013 #6

    Mark44

    Staff: Mentor

    I suppose, but why would you not want to use a graph? That's probably the simplest way to proceed.
     
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