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    Find the solution of this equation:

    Cot(x)= 3Sin2(x) interval -180<X<180

    i got the worked solution, however its not making very much sense and I want to understand this.

    Thanks
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    (Original post by Stanza)
    3Sin2(x)
    is that

    3sin2x

    or

    3sin^2x

    ?
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    (Original post by Pheylan)
    is that

    3sin2x

    or

    3sin^2x

    ?
    its the first one
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    its c2!

    Heres an identity:
    sin2(x)+cos2(x)=1

    re-arrange:
    sin2(x)=1-cos2(x)
    sub in for sin2(x)

    cos(x)=3(cos2(x))

    then its easy
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    oh.

    wow i read that wrong twice

    cot (x)

    also sin (2x)

    sorry dude
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    (Original post by Stanza)
    its the first one
    cotx = 3sin2x

    \frac {cosx}{sinx} = 3(2sinxcosx)

    cosx = 6cosxsin^2x

    cosx = 6cosx(1 - cos^2x)

    cosx = 6cosx - 6cos^3x

    6cos^3x - 5cosx = 0

    cosx(6cos^2x - 5) = 0

    cosx = 0

    cos^2x = \frac {5}{6}

    cosx = \pm \sqrt \frac {5}{6}
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    (Original post by Jizkid Jnr)
    its c2!

    Heres an identity:
    sin2(x)+cos2(x)=1

    re-arrange:
    sin2(x)=1-cos2(x)
    sub in for sin2(x)

    cos(x)=3(cos2(x))

    then its easy
    I think there is a misunderstanding. if I write it like this, maybe its better
    cotØ=3sin2Ø and there is no squared, its a normal 2
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    (Original post by Stanza)
    I think there is a misunderstanding. if I write it like this, maybe its better
    cotØ=3sin2Ø and there is no squared, its a normal 2
    read my post
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    yes thanks
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    (Original post by Stanza)
    yes thanks
    but the answer in the book says: square root of 1/6???????????????????
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    (Original post by Stanza)
    but the answer in the book says: square root of 1/6???????????????????
    Well sin(x) = square root of 1/6


    3*(2sin(x)cos(x))= cos(x) / sin(x)

    = 6 sin^2(x)cos(x) = cos(x)

    = 6sin^2(x)=1 (dividing both sides by cos(x) )

    =sin^2(x)= 1/6

    sin(x) = root (1/6)
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    (Original post by Stanza)
    Find the solution of this equation:

    Cot(x)= 3Sin2(x) interval -180<X<180

    i got the worked solution, however its not making very much sense and I want to understand this.

    Thanks
    cotx=3sin2x

    1) write cotx as : cosx/sinx
    2) And you know that sin2x is the same as 2sinxcos

    therefore you have : cosx/sinx=3(2sinxcosx)
    3) expand the rhs -- cosx/sinx=6sinxcosx
    4)multiply both sides by sinx....this gives :

    cosx=6sin^xcosx

    5) now make it equal zero....

    6sin^xcosx-cosx=0

    6)But this can be facored since cosx is common to both
    7)so.... cosx(6sin^x-1)=0


    now its really easy.... cosx=0
    0r 6sin^x-1=0

    therefore arcos0=90

    arcos(+/- .. (1/6))= 0

    so you do arcos both plus and minus 1/6

    remember that cos(360+/-x)=cosx
    and sin(180-x)=sinx

    is that correct>? or have i left something out?
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    (Original post by sabuuk)
    is that correct>?
    no
    (Original post by sabuuk)
    have i left something out?
    yes, the correct answer

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    (Original post by Pheylan)
    no

    yes, the correct answer

    lol...but he didn't want the answer, he just wanted some understanding
 
 
 
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