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    Heya! i'm just starting my new trig section
    sorry if its not so clearly set out, as it's the second time i write this post. can't be bothered with latexing it all this time sorry
    I''m stuck with this guys, so if could give me a nudge in the right dirrection i'd appreciate it.

    Q:cos2x=cos(x+pi/3) find general soln
    i did:
    RHS: cosxcospi/3-sinxsinpi/3
    = 1/2cosx-sqrt3/2sinx

    LHS:cos^2x-sin^2x
    =>
    cos^2x-sin^2x=1/2cosx-sqrt3/2sinx

    hence  cos^2x-1/2cosx+sqrt3/2sinx-sin^2x=0

    so (cosx-1/4)^2-1/16+(-sinx+sqrt3/4)^2-3/16=0

    thus cosx=1/4+-sqrt(1/16) or -sinx=(-sqrt3/4+-sqrt(3/16))

    giving general solutions of x=npi and x=npi/2+(-1)^n.0
    which are not correct.
    again i apologise for the quickness of writing this, as it was so nice and pretty before :'(
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    (Original post by Goldenratio)
    Heya! i'm just starting my new trig section
    sorry if its not so clearly set out, as it's the second time i write this post. can't be bothered with latexing it all this time sorry
    I''m stuck with this guys, so if could give me a nudge in the right dirrection i'd appreciate it.

    Q:cos2x=cos(x+pi/3) find general soln
    i did:
    RHS: cosxcospi/3-sinxsinpi/3
    = 1/2cosx-sqrt3/2sinx

    LHS:cos^2x-sin^2x
    =>
    cos^2x-sin^2x=1/2cosx-sqrt3/2sinx

    hence  cos^2x-1/2cosx+sqrt3/2sinx-sin^2x=0

    so (cosx-1/4)^2-1/16+(-sinx+sqrt3/4)^2-3/16=0

    thus cosx=1/4+-sqrt(1/16) or -sinx=(-sqrt3/4+-sqrt(3/16))

    giving general solutions of x=npi and x=npi/2+(-1)^n.0
    which are not correct.
    again i apologise for the quickness of writing this, as it was so nice and pretty before :'(
    some general solutions are;
    x+ π/3 = 2nπ - 2x
    => 3x = 2nπ - π/3
    => x = 2nπ/3 - π/9

    2nπ - (x+π/3) = 2x
    2nπ - x - π/3 = 2x
    2nπ - π/3 = 3x
    => x = 2nπ/3 - π/9 <= (same result as above)

    2x = x + π/3 + 2nπ
    => x = π/3 + 2nπ

    x+ π/3 = 2nπ + 2x
    => x = π/3 - 2nπ
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    (Original post by Goldenratio)
    Q:cos2x=cos(x+pi/3) find general soln
    (
    If STEP taught me one thing, it was the whole "cosx = cosy doens't imply that x=y" argument. Then again, I undoubtedly failed STEP miserably this year so I'll give you a haphazard solution which doesn't feel right :

    Considering the cos graph: The period of the graph y = cosx is 2π (i.e. How long it takes before the graph begins to 'loop'), so a general solution will have a multiple of 2pπ in it as a result. In other words, cosx = cosy => x = y + 2pπ

    cos2x = cos(x + π/3)

    => 2x = (x + π/3) + 2pπ (p is included in the set of integers)

    x = π[1/3 + 2p]

    x = (π/3)[1 + 6p]

    I reckon it's wrong though .
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    You can use the general result that if cosA=cosB, B=2npi-A,or B=2npi+A,nEZ.
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    \cos 2x - \cos (x+\frac{\pi}{3}) = 0 \\

\cos (A+B) - \cos (A-B)=-2sinAsinB \\

A+B=2x, A-B=x+\frac{\pi}{3} \\

A=\frac{3x}{2}+\frac{\pi}{6} \\

B=\frac{x}{2}-\frac{\pi}{6} \\

-2\sin (\frac{3x}{2}+\frac{\pi}{6} ) \sin (\frac{x}{2}-\frac{\pi}{6} ) = 0 \\

\sin (\frac{x}{2}-\frac{\pi}{6} ) \sin (\frac{3x}{2}+\frac{\pi}{6})=0 \\

\frac{x}{2}-\frac{\pi}{6}=n\pi \textit{ or } \frac{3x}{2}+\frac{\pi}{6}=n\pi \\

x=2n\pi + \frac{\pi}{3} \textit{ or } x=\frac{2}{3}n\pi - \frac{\pi}{9}
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    Hey! Thank you guys very much for all your help! Wow that makes my sunday morning!
 
 
 
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