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Factoring with cosine

  1. Nov 11, 2013 #1
    1. The problem statement, all variables and given/known data

    9cos(2t)=9cos2(t)-4

    for all the smallest four positive solutions
    2. Relevant equations



    3. The attempt at a solution

    I've factored it and pulled out 9cos(t) and made that = u
    so i have u2-2u-4
    That factors to 3.23606 and -1.236

    Next i added 9cos(t) back in. 9cos(t)=3.236 and 9cos(t)=-1.236
    then solve for the values for the inverse cos-1(3.236/9) and cos-1(-1.236/9)



    am i on the right track? I'm not entirely sure if i'm doing it correctly
     
  2. jcsd
  3. Nov 12, 2013 #2

    Mark44

    Staff: Mentor

    No. You are thinking the equation is quadratic in form - it isn't. Use the double-angle identity to rewrite cos(2t) in a different form.
     
  4. Nov 12, 2013 #3
    ak ok now i have 8cos2(t)-9sin2(t)+4

    is this better?
     
  5. Nov 12, 2013 #4

    Dick

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    How did you get that??
     
  6. Nov 12, 2013 #5
    ok so i used the double angle forumla cos(2t)=cos2(t)-sin2(t)
    9(cos2(t)-sin2(t))=cos2(t)-4
    distributed the 9 and 9cos2(t)-9sin2(t))=cos2(t)-4

    then subtracted cos2 and added 4. 8cos2(t)-9sin2(t)+4

    I'm not doing something right...
    I'm thinking i need the half angle formula cos2=1+cos(2u)/2
     
  7. Nov 12, 2013 #6
    whoops messed something up here
     
  8. Nov 12, 2013 #7

    Dick

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    The original equation you posted was 9cos(2t)=9cos2(t)-4. What happened to the second 9? Or was that a typo?
     
  9. Nov 12, 2013 #8
    typo. the 9cos2(t) cancelled out now i have -9sin2=-4 -> sin2=4/9
     
  10. Nov 12, 2013 #9

    Dick

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    That's better.
     
  11. Nov 12, 2013 #10
    now ive squarerooted it and now have sin(t)=2/3 now to arcsin(2/3)=.7297. That is one solution

    ( i think)

    then i think i subtract 2pi. 2pi-.7297=5.5534 (second solution i think). I add and subtract pi to .7297 because 2pi would make it too big

    so pi-.7297=2.4118
    pi+.72972=3.871
     
  12. Nov 12, 2013 #11

    Dick

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    Sounds about right. But your logic is a little fuzzy. sin(t)=(-2/3) is also a solution. Graph y=sin(t), y=2/3 and y=(-2/3) in the range [0,2*pi] to make sure you understand why all of those work.
     
    Last edited: Nov 12, 2013
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