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    It's from Pearson's Advanced Maths A2 Core for Edexcel
    pg65 Ex3C Q8b

    Solve for values of x between 0° and 360°

    cos(x + 45°) = cosx

    I've tried to work it out but I get stuck and we didn't do a question similar to this in class to compare it to

    This is my working so far but it doesn't leave me much to go off so I presume it's incorrect.
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    (Original post by JimJam707)
    ...
    You initially had a "cos x" on the LHS (correctly), but crossed it out for some reason and treated it as zero. Three lines up.
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    (Original post by ghostwalker)
    You initially had a "cos x" on the LHS (correctly), but crossed it out for some reason and treated it as zero. Three lines up.
    Adding back the cosx only leads me up to
     2cosx = \sqrt{2}cosx - \sqrt{2}sinx
    and then I am stuck again
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    (Original post by JimJam707)
    Adding back the cosx only leads me up to
     2cosx = \sqrt{2}cosx - \sqrt{2}sinx
    and then I am stuck again
    collect the cos elements on one side and the sin elements on the other side

    then rearrange to get tan(x) = something
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    (Original post by JimJam707)
    Adding back the cosx only leads me up to
     2cosx = \sqrt{2}cosx - \sqrt{2}sinx
    and then I am stuck again
    So (2-\sqrt2)\cos x+\sqrt2 \sin x=0 Now just divide by \cos x
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    (Original post by JimJam707)
    Adding back the cosx only leads me up to
     2cosx = \sqrt{2}cosx - \sqrt{2}sinx
    and then I am stuck again
     2cosx - \sqrt{2}cosx = - \sqrt{2}sinx

     (2 - \sqrt{2})cosx = - \sqrt{2}sinx

     (2 - \sqrt{2}) = - \sqrt{2} sinx / cosx

     (2 - \sqrt{2}) / \sqrt{2} = -  sinx / cosx

      tanx = - (2 /\sqrt{2} - \sqrt{2} / \sqrt{2})

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    (Original post by Tutorsos)

    The forum guidelines ask us to hint and guide - NOT to give full solutions
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    (Original post by JimJam707)
    It's from Pearson's Advanced Maths A2 Core for Edexcel
    pg65 Ex3C Q8b

    Solve for values of x between 0° and 360°

    cos(x + 45°) = cosx

    I've tried to work it out but I get stuck and we didn't do a question similar to this in class to compare it to

    This is my working so far but it doesn't leave me much to go off so I presume it's incorrect.
    I think the simplest way to use that rule when

    \displaystyle \cos A =\cos B

    then the 2 cases are:

    1: \displaystyle A=B + 2k\pi
    or
    2: \displaystyle A=-B + 2k\pi
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    (Original post by ztibor)
    I think the simplest way to use that rule when

    \displaystyle \cos A =\cos B

    then the 2 cases are:

    1: \displaystyle A=B + 2k\pi
    or
    2: \displaystyle A=-B + 2k\pi
    or A=B?
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    (Original post by TenOfThem)
    or A=B?
    ???
 
 
 
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Updated: September 23, 2014
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