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    Same value of x as last 2 threads i made

    tanx=cosx

    sinx=cos²x

    sinx=1-sin²x

    sin²x+sinx-1=0

    let sinx=y

    y²+y-1=0

    quaratic equation gives me (-1±root 2)/2

    then i got a wrong answer >.>
    help, where have i made a stupid mistake this time

    TeeEm Zacken
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    (Original post by thefatone)
    Same value of x as last 2 threads i made

    tanx=cosx

    sinx=cos²x

    sinx=1-sin²x

    sin²x+sinx-1=0

    let sinx=y

    y²+y-1=0

    quaratic equation gives me (-1±root 2)/2






    then i got a wrong answer >.>
    help, where have i made a stupid mistake this time

    TeeEm Zacken
    sorry I am teaching
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    (Original post by TeeEm)
    sorry I am teaching
    oh.. ok
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    (Original post by thefatone)
    Same value of x as last 2 threads i made

    tanx=cosx

    sinx=cos²x

    sinx=1-sin²x

    sin²x+sinx-1=0

    let sinx=y

    y²+y-1=0

    quaratic equation gives me (-1±root 2)/2

    then i got a wrong answer >.>
    help, where have i made a stupid mistake this time

    TeeEm Zacken
    There's where.
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    (Original post by thefatone)
    Same value of x as last 2 threads i made

    tanx=cosx

    sinx=cos²x

    sinx=1-sin²x

    sin²x+sinx-1=0

    let sinx=y

    y²+y-1=0

    quaratic equation gives me (-1±root 2)/2

    then i got a wrong answer >.>
    help, where have i made a stupid mistake this time

    TeeEm Zacken
    As above, solving the quadratic equation gives:

    \displaystyle y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2 \times 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \cdots
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    1 + 4 = 5 not 2
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    (Original post by Zacken)
    As above, solving the quadratic equation gives:

    \displaystyle y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2 \times 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \cdots
    (Original post by the bear)
    1 + 4 = 5 not 2
    thanks guys so much, i was being stupid again >.>
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    (Original post by the bear)
    1 + 4 = 5 not 2
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    (Original post by Zacken)
    claiming this gif for future use
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    (Original post by Zacken)
 
 
 
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