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Math Help - factoring a trig function

  1. #1
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    factoring a trig function

    I can do the basic ones, but when they have a coefficient in front of the first term I dont know what to do.
    2cos^2x-7cosx-3

    I thought maybe you do

    (2cosx-1)(cosx-3)
    Edit nevermind That is right.
    BUt is there a better way to solve these other than putting the first term in the first bracket and the guessing at multiples of the last term?
    because with one like 6sin^2x+sinx-1 I get (6sinx-1)(sinx+1)
    and its supposed to be(2sinx-1)(3sinx-1)

    Im confused with factoring these, could I have some help please.
    thanks.
    Last edited by brentwoodbc; May 2nd 2009 at 01:50 PM.
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  2. #2
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    Quote Originally Posted by brentwoodbc View Post
    I can do the basic ones, but when they have a coefficient in front of the first term I dont know what to do.
    2cos^2x-7cosx-3

    I thought maybe you do

    (2cosx-1)(cosx-3)
    but that doesnt work.


    Im confused with factoring these, could I have some help please.
    thanks.
    You can factorise it like a normal quadratic: Let u = cos(x) :

    2u^2 - 7u - 3 = 0

    Since the discriminant does not equal a perfect square this cannot be factorised rationally. You may solve it using the quadratic formula or completing the square:

    cos(x) = u = \frac{-b\pm \sqrt{b^2-4ac}}{2a}
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    You can factorise it like a normal quadratic: Let u = cos(x) :

    2u^2 - 7u - 3 = 0

    Since the discriminant does not equal a perfect square this cannot be factorised rationally. You may solve it using the quadratic formula or completing the square:

    cos(x) = u = \frac{-b\pm \sqrt{b^2-4ac}}{2a}
    its 2u^2 - 7u + 3 = 0 btw sorry
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  4. #4
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    Quote Originally Posted by brentwoodbc View Post
    thanks so when I find x=... how do I give my answer in factor form, he answer given is(2cosx-1)(cosx-3)
    (My last post is only valid if the equation is equal to 0)
    Do you mean

    <br />
2cos^2x-7cosx-3<br />
or

    <br />
2cos^2x-7cosx+3<br />
?

    Since the second one factors I will assume you meant that one:

    Let u = cos(x) : 2u^2-7u+3

    This would equal (2u )(u ). The other numbers will be 1 and 3 and there will be two minus signs (because the last term is positive). The sum of 2 times one of them plus the other will be equal to -7.

    Spoiler:
    (2u-1)(u-3)
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  5. #5
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    Smile

    Quote Originally Posted by e^(i*pi) View Post
    (My last post is only valid if the equation is equal to 0)
    Do you mean

    <br />
2cos^2x-7cosx-3<br />
or

    <br />
2cos^2x-7cosx+3<br />
?

    Since the second one factors I will assume you meant that one:

    Let u = cos(x) : 2u^2-7u+3

    This would equal (2u )(u ). The other numbers will be 1 and 3 and there will be two minus signs (because the last term is positive). The sum of 2 times one of them plus the other will be equal to -7.

    Spoiler:
    (2u-1)(u-3)
    thanks ya I made a mistake it is +3
    Thank you. Im a bit rusty a factoring.
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  6. #6
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    Quote Originally Posted by brentwoodbc View Post
    thanks ya I made a mistake it is +3
    Thank you. Im a bit rusty a factoring.
    I forgot to mention it in my last post but don't forget to change u back to cos(x)
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