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Find interval of X

  1. Aug 7, 2014 #1
    1. The problem statement, all variables and given/known data

    2sin2x + 3sinx - 2>0 and x2 - x - 2 < 0 . x lies in the interval

    A)∏/6 to 5∏6
    B)-1 to 5∏/6
    C)-1 to 2
    D) ∏/6 to 2

    Correct Answer is D, and I get C.

    2. Relevant equations



    3. The attempt at a solution

    So I factorized the second equation and obtained (x+1)(x-2)<0
    So this, on drawing a number line, shows that x lies between -1 and 2, both excluded.

    Next, I factorized the first equation, which gave me (2sinx-1)(sinx+2)>0
    Plotting on a number line, sin x is either less than -2 or greater than 1/2. Former is not possible.
    That indicates that x lies between ∏/6 (=approx 0.523) and 5∏/6 (= approx 2.61), and all periodic repetitions of those two angles, as sin x is only greater than 1/2 for those values in between.

    So the intersection of the first equation's number line( from -1 to 2) and this equations number line( from 0.523 to 2.61), gives me -1 to 2. This is apparently wrong, according to the answer in the book, which is ∏/6 to 2.

    I'm rather tired, so I can't wrap my head around any stupid mistakes. Please do help :)
     
    Last edited: Aug 7, 2014
  2. jcsd
  3. Aug 7, 2014 #2

    LCKurtz

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    Correct so far.

    Also correct.

    No, that gives you ##(\frac \pi 6, 2)##. The values of ##x## must be in both solution sets so anything less than ##\frac \pi 6## or greater than ##2## doesn't work.
     
  4. Aug 7, 2014 #3
    Wow. That was dumb of me.
    Thanks so much.
     
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