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    Hi, please can someone help me solve the following equation in the interval 0<x<2pi

    Cosx(sinx-3cosx)=0

    Thanks
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    Posted from TSR Mobile

    Multiply out the brackets?
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    (Original post by Hlrk)
    Hi, please can someone help me solve the following equation in the interval 0<x<2pi

    Cosx(sinx-3cosx)=0

    Thanks
    What do you get after that...

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    Because it's equal to zero you can solve cosx=0 and (sinx-3cosx)=0 think why this holds true.

    Also this thread should be in the maths forum
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    I've multiplied out the brackets and I've managed to get tanx=3cosx. Not sure where to go from there
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    (Original post by Hlrk)
    I've multiplied out the brackets and I've managed to get tanx=3cosx. Not sure where to go from there
    Firstly ... ... Do not multiply out the brackets

    You have a factorised form a(b-c)=0

    So, either a=0 or b-c=0
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    (Original post by ttaylor17)
    Posted from TSR Mobile

    Multiply out the brackets?
    There's no need to multiply out the brackets - it would make things less clear here!

    If AB = 0 then you know that one (or both) of A and B is 0.
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    So do I just do cos^-1(0)?
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    (Original post by Hlrk)
    So do I just do cos^-1(0)?
    That would give two of your answers

    The bracket gives the others
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    (Original post by Hlrk)
    Hi, please can someone help me solve the following equation in the interval 0<x<2pi

    Cosx(sinx-3cosx)=0

    Thanks
    cosx=0 and sinx=3cosx => tanx =1/3
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    (Original post by WatermelonJuice1)
    cosx=0 and sinx=3cosx => tanx =1/3
    Er, sin x = 3cos x => tan x = 3
 
 
 
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