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Inequalities find the set of values of x

  1. Oct 25, 2012 #1
    find the set of values of x for which [tex] 2x > \dfrac{3x + 1}{x+1} [/tex] done this and got [tex] -1<x<-0.5, x > 1 [/tex]

    b) find the set of inequalities where

    [tex] 2sint > \dfrac{3sint + 1}{sint + 1} [/tex] where [tex] -\pi < t < \pi [/tex]

    first I found the set values of t suitable in the range for -1,-0.5,1 which I got to be as [tex] -\frac{\pi}{2}, - \frac{\pi}{6}, - \frac{5\pi}{6}, \frac{pi}{2} [/tex] and hence getting [tex] \frac{-5\pi}{6} < t < \frac{-\pi}{6}, t > \frac{\pi}{2} [/tex]

    however I'm not sure if it is correct, and if it isn't I don't know how else to do it.
     
    Last edited: Oct 25, 2012
  2. jcsd
  3. Oct 25, 2012 #2

    SammyS

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    You did the first part correctly; the part where you solved for x.

    You can directly substitute sin(t) for x in that result.

    [itex]\displaystyle -1<\sin(t)<-0.5\,,\ \sin(t)>1[/itex]

    I sin(t) ever greater than 1 ?
     
  4. Oct 25, 2012 #3
    No it's not, also are the inequalities supposed to be in terms of t?
     
  5. Oct 25, 2012 #4

    SammyS

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    Yes. Your final answer should be in terms of t, but you can eliminate sin(t) > 1, from contributing anything to the solution.
     
  6. Oct 25, 2012 #5
    [tex] \frac{-5\pi}{6} < t < \frac{-\pi}{6} [/tex]

    is that the correct answer then?
     
  7. Oct 25, 2012 #6

    SammyS

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    Looks good !
     
  8. Oct 25, 2012 #7
    Thanks!
     
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