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Math Help - Help with trig and radians

  1. #1
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    Help with trig and radians

    Fidn all the values of θ with θ_< θ< 2pi such that sin(3θ)= -0.5
    Last edited by mr fantastic; July 12th 2010 at 07:34 PM. Reason: Deleted excessive ?'s in post title.
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  2. #2
    A Plied Mathematician
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    Two questions:

    1. Is that 0<\theta<2\pi?
    2. What have you done so far?
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  3. #3
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    1. Yep that's right.
    2. I haven't done anything yet.
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  4. #4
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    Ok. You need to show the initiative on this forum. We'll help you get unstuck, but the rule is that people asking for help basically have to do the work. Otherwise, we're not helping you understand. So, show us some work, and we'll help you get unstuck.
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  5. #5
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    I honestly have no clue what to do, or where to start from to solve this problem.
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  6. #6
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    Ok. Try plotting the function over the region indicated, and see where that leads you.
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  7. #7
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    What do you mean by that :S
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  8. #8
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    Plotting, graphing, whatever. Use a calculator or a computer to show a graph of the function over the interval from 0 to 2\pi.
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  9. #9
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    I haven't got a calculator or computer program that can do this.
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  10. #10
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    Ok. Go to Wolfram Alpha and try this on for size. That's obviously not the function you want, but you can get what you want. Trig functions must be typed as follows: cos(x) is typed as Cos[x].
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  11. #11
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    Exclamation

    Hi,

    It's a simple Trig equation and the problem wants you to find all the angles in that interval such that the equation holds for them. It can simply be solved and you don't need to graph it by calculator!!!
    ----------------------------
    sin(3x)=-.5=sin(-pi/6) => 3x=2kpi-pi/6 and 3x=2kpi+pi-(-pi/6). => 3x=2kpi+7pi/6
    since 0<=x<2pi => 0<=3x<6pi. Now we want to fint all the ks such that the last inequality holds. By getting k=0 and k=1, that holds! that is if k=0 => x=7pi/6 . if k=1 => x=11pi/6. The equation has another answers for those ks; but the red ones are the only acceptable solutions!
    **********************
    Note: you could calculate the solution sets for x and then find xs in the first interval; but may be a bit hard!
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