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    Hence or otherwise show that the equation:

    cosx (cos2x/cosx + sinx) = 1/2, can be written as:

    sin2x = cos2x


    afterwards, how do I re-arrange to solve 'sin2x = cos2x', 0< x < 2pi
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    clearly sin 2x =0 or cos 2x=0 are not solutions. so we can divide by one of them.
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    (Original post by W.H.T)
    Hence or otherwise show that the equation:

    cosx (cos2x/cosx + sinx) = 1/2, can be written as:

    sin2x = cos2x


    afterwards, how do I re-arrange to solve 'sin2x = cos2x', 0< x < 2pi
    Divide both sides by cos2x to get tan2x =1

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    (Original post by W.H.T)
    Hence or otherwise show that the equation:

    cosx (cos2x/cosx + sinx) = 1/2, can be written as:

    sin2x = cos2x

    and afterwards, how do I re-arrange to solve 'sin2x = cos2x', 0< x < 2pi
    multiply out the brackets...
    cos2x + sinx cosx = 1/2

    then use this identity...
    sin(2x) = 2 sinx cosx

    and then to solve divide both sides by cos(2x), so sin(2x)/cos(2x) = tan(2x) = 1.
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    (Original post by kat_s)
    multiply out the brackets...
    cos2x + sinx cosx = 1/2

    then use this identity...
    sin(2x) = 2 sinx cosx

    and then to solve divide both sides by cos(2x), so sin(2x)/cos(2x) = tan(2x) = 1.

    for the first step of multiplying out the brackets, would I get:

    cos^2 2x / cosx+sinx = 1/2
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    (Original post by Superman_Jr)
    Divide both sides by cos2x to get tan2x =1

    sin2x = cos2x

    how can I divide by cos2x, because on the 'cos2x' side there is nothing multiplying it. So surely to get 'cos2x' to the other side, I would have to subtract by 'cos2x' ?
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    (Original post by W.H.T)
    sin2x = cos2x

    how can I divide by cos2x, because on the 'cos2x' side there is nothing multiplying it. So surely to get 'cos2x' to the other side, I would have to subtract by 'cos2x' ?
    when you divide by cos2x, you'd be left with 1. And you can divide by cos2x as long as you do the same to both sides, the equation remains unchanged.
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    (Original post by W.H.T)
    for the first step of multiplying out the brackets, would I get:

    cos^2 2x / cosx+sinx = 1/2
    cosx (cos2x/cosx + sinx) = 1/2

    (cosx cos2x)/cosx + cosx sinx = 1/2

    the cos x at the top and bottom of the first fraction cancel.

    Unless you mean cosx (cos2x/(cosx +sinx)) for the original equation??
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    (Original post by kat_s)
    cosx (cos2x/cosx + sinx) = 1/2

    (cosx cos2x)/cosx + cosx sinx = 1/2

    the cos x at the top and bottom of the first fraction cancel.

    Unless you mean cosx (cos2x/(cosx +sinx)) for the original equation??

    yeah my original equation is: cosx (cos2x/(cosx +sinx)) = 1/2

    I don;t understand how you got the part in bold here:
    (cosx cos2x)/cosx + cosx sinx = 1/2
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    how do you solve 2sin(2x)+4cos^2(x)=3
 
 
 
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