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    Especially where a and b are non integers so you can't use identities.

    You can't just inverse sine them since you either get the trivial x=0 or a contradiction.

    Also the same for cos(ax)=cos(bx) and tan(ax)=tan(bx).
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    cos(ax)=cos(bx)

    for this identity a and b would be equall
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    Subtract one of the terms from both sides.
    You get sin(ax)-sin(bx)=0
    Use identity sinA - sinB =2sin[(A-B)/2]cos[(a+b)/2]
    In this case A=ax and B=bx
    Then either sin[(a-b)x/2]=0 or cos[(a+b)x/2]=0
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    (Original post by Katiee224)
    cos(ax)=cos(bx)

    for this identity a and b would be equall
    Are you sure?

    Plotting cos(x/2) and cos(2x/3) shows a solution at 12pi/7 .
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    (Original post by Katiee224)
    cos(ax)=cos(bx)

    for this identity a and b would be equall
    a =b, what about a=-b, a=b+2π.....
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    (Original post by Ano123)
    Subtract one of the terms from both sides.
    You get sin(ax)=-sinbx=0
    Use identity sinA - sinB =2sin[(A-B)/2]cos[(a+b)/2]
    In this case A=ax and B=bx
    Then either sin[(a-b)x/2]=0 or cos[(a+b)x/2]=0
    Oh thanks that's a great one, I almost never use these identities so it's easy to forget. Does one exist for tan?
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    (Original post by 16characterlimit)
    Oh thanks that's a great one, I almost never use these identities so it's easy to forget. Does one exist for tan?

    Convert tan to expression with sin and cos and then combine fractions.
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    (Original post by Ano123)
    Convert tan to expression with sin and cos and then combine fractions.
    Thanks.
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    It might be interesting if you try to find the general solution to the equation in terms of and b.
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    Not sure what the above answers are trying to achieve. The general solution to \sin x = \sin y is x = y + 2k \pi or x = (2k+1)\pi - y.

    So substituting in x \mapsto ax and y \mapsto bx gets you what you need.
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    I am not great at basic mathematics, but I ave some understanding of the more advanced fields of maths.

    If I understand the question correctly,

    sin(ax)=sin(bx) tan(ax)=tan(bx) cos(ax)=cos(bx) the answer I think is

    sin(ab)=sin(2x) tan(ab)=tan(2x) cos(ab)=cos(2x)

    can someone tell me if I am right on this, as I would be well chuffed if I am.
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    (Original post by Zacken)
    Not sure what the above answers are trying to achieve. The general solution to \sin x = \sin y is x = y + 2k \pi or x = (2k+1)\pi - y.

    So substituting in x \mapsto ax and y \mapsto bx gets you what you need.
    I agree that this is the neatest approach. However, you can also use a factor formula method as suggested earlier.
 
 
 
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