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Math Help - Solving Trigonometry Equations Cosine/Sine graph help

  1. #1
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    Solving Trigonometry Equations Cosine/Sine graph help

    I am really stuck in this question. if someone could help me, that'd be awesome.

    Solve the equation 2cos^2x + cosx-3=0, where -pi is less than or equal to x, which is less than or equal to +pi. Then make a general solution for that equation.

    I am just so clueless with math in general.
    Even factoring the equation is confusing me.
    Does 2cos^2x +cosx -3 Factor out to be: (2cosx+3)(Cosx-2) or does it factor out to be (cosx-1)(Cosx+3)

    Because depending on whether you put the "-2" in the same brackets as the Coefficient of 2 (Ie. [2cosx-2] or [2cosx-3]) It will factor out differently.

    If someone could help me with the whole equation in general. that'd help. thank you!
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  2. #2
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    Quote Originally Posted by Radiofruit View Post


    Does 2cos^2x +cosx -3 Factor out to be: (2cosx+3)(Cosx-2) or does it factor out to be (cosx-1)(Cosx+3)
    neither

    in your case use the quadratic formula

    \cos(x) = \frac{-1\pm\sqrt{1^2-4\times 2\times -3}}{2\times 2}
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  3. #3
    Super Member bigwave's Avatar
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    2cos^2x + cosx-3=0

    (2cosx +3)(cosx-1)=0

    cos^{-1}(1) = 0

    2cos^2(0) + cos(0)-3=0
    Last edited by bigwave; January 6th 2010 at 06:11 PM.
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  4. #4
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    hmm

    How do you know that you're suppose to use the quadratic formula as opposed to factoring? is there a way to know?
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  5. #5
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    You can always use the quadratic formula on a quadratic equation, sometimes it is difficult to factor them straight out as you have experienced. bigwave's solution is excellent but sometimes a bit tricky and time consuming.

    The real difference is not alot, the quadratic formula solves the equation straight away where factoring is just a step in the solving process.
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  6. #6
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    Question part 3

    The thing that really just confuses me is
    if you use the factoring method, how do you know whether you're suppose to factor (2cosx-3) or (2cosx-2)--->(cosx-1)

    How do you figure out which one's the right way to factor it?
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  7. #7
    Super Member bigwave's Avatar
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    well agreeing with pickslides this is where the quadradic formula would be best

    however this did factor

    set both factors to 0 and solve
    in this case on (cos(x)- 1) would solve the other was undefined
    ussually tho all factors will solve.

    as mentioned factoring can be tricky I just happen to see it off hand
    which ussually is not the case when one is has not done it a lot

    hope i didn't cause any confusion...
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  8. #8
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    Quote Originally Posted by Radiofruit View Post
    The thing that really just confuses me is
    if you use the factoring method, how do you know whether you're suppose to factor (2cosx-3) or (2cosx-2)--->(cosx-1)
    They are factors.


    Quote Originally Posted by Radiofruit View Post
    (2cosx-3) or (2cosx-2)
    these 2 are incorrect

    Quote Originally Posted by Radiofruit View Post
    (cosx-1)
    this one is correct, I think you are asking how do you know which ones are correct?

    Read this to get some backgorund

    Cross-Multiplication Method
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  9. #9
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    Thank you!

    Okay, I think i've got it! thank you guys.
    I am embarassingly bad at math.
    Gah.
    Thanks for both of your help, I think i'm good now.
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