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    When cos4\theta = cos 3\theta, prove that \theta = 0, 2\pi/7, 4\pi/7 , 6\pi/7.

    This could be done by considering the general solution but it wouldn't really be a proof. I could do this:

    \displaystyle \frac{1}{2}(e^{4\theta i} + e^{-4\theta i}) = \frac{1}{2}(e^{3\theta i} + e^{-3\theta i}) \Rightarrow e^{8\theta i } + 1 = e^{7\theta i} + e^{\theta i}

 \Rightarrow e^{7\theta i}(e^{\theta i } - 1) - (e^{\theta i } - 1) = 0

    So we have e^{\theta i} = 1, e^{7 \theta i} = 1 which will give us the required solutions.

    However, is there an alternative and simpler proof for this? I ask because this question comes right after introduction to de Moivre's theorem way before hyperbolic functions.
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    (Original post by Ateo)
    When cos4\theta = cos 3\theta, prove that \theta = 0, 2\pi/7, 4\pi/7 , 6\pi/7.

    This could be done by considering the general solution but it wouldn't really be a proof. I could do this:

    \displaystyle \frac{1}{2}(e^{4\theta i} + e^{-4\theta i}) = \frac{1}{2}(e^{3\theta i} + e^{-3\theta i}) \Rightarrow e^{8\theta i } + 1 = e^{7\theta i} + e^{\theta i}

 \Rightarrow e^{7\theta i}(e^{\theta i } - 1) - (e^{\theta i } - 1) = 0

    So we have e^{\theta i} = 1, e^{7 \theta i} = 1 which will give us the required solutions.

    However, is there an alternative and simpler proof for this? I ask because this question comes right after introduction to de Moivre's theorem way before hyperbolic functions.
    Sorry but I fail to see why considering the general solution would not be acceptable. It is really no different to what you have done with the complex number solution.

    4\theta = \pm 3\theta + 2n\pi
    Either 7\theta = 2n\pi or \theta = 2n\pi.

    It doesn't get much simpler than that.


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    You could write cos(4t)=cos(3t) as functions of cos(t) then solve that.
 
 
 
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