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Trig Inequality

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Given 0≤x<2pi, solve tan x/2 > -1



    3. The attempt at a solution
    I thought I would set tan x/2=-1 but I'm not sure.

    Is the answer 0<x<pi, 3pi/2<x<2pi?
     
    Last edited: Feb 6, 2009
  2. jcsd
  3. Feb 6, 2009 #2
    Your answer doesn't look right to me.

    Why not try a graphical approach? From 0 to 2pi there will be 4 periods, and it looks like there will be 4 intervals in your solution.

    Oops! The period is 2pi, not pi/2. Yes, it seems your solution is right. Without a graph, I like your approach of solving the equality. Then plug in some test points on either side of each solution you get. Pay attention to asymptotes (x = pi in this case).
     
  4. Feb 6, 2009 #3

    HallsofIvy

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    Yes, solving tan(x/2)= -1 is the way to start. On any interval NOT including a root of tan(x/2)= -1 and NOT including [itex]\pi[/itex], where tan(x/2) is not continuous, tan(x/2) is always less than -1 or always less than -1. Determine what those intervals are and check one point in each interval to see if tan(x/2), on that interval, is greater than -1.
     
  5. Feb 9, 2009 #4
    I got one question.

    What if I have cosx > -1 and 0≤x<2п ?

    x Є (-п, п) or x Є (п,3п) оr x Є (п,п+2kп), k Є Z

    But because of 0≤x<2п would x Є (0,п) or x Є (п,2п) or x Є ( kп, п(k+1) ) where k Є Z.

    Am I right?
     
  6. Feb 9, 2009 #5
    Your first answer is closer than your second answer. If you're restricted to [0, 2pi) then there's no need to mess with the k values.

    Two hints: cos(0) = 1 and cos(pi) = -1
     
  7. Feb 10, 2009 #6
    Yes, you're right. I don't want to mess with k values.

    So my answer is (0,п) or (п,2п)

    And what about cosx < -1 . As I know there isn't smallest value than -1 in cos. So is x Є empty set?

    Thanks for the help.

    Regards.
     
  8. Feb 10, 2009 #7

    HallsofIvy

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    Yes, {x| cos(x)< -1} is the empty set.
     
  9. Feb 10, 2009 #8
    Right. And remember that, for the original problem, zero is part of the solution. So your interval should start with a [ instead of a (
     
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