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I am trying to solve the equation $$\tan(2x) = 1+\tan(x).$$

I have tried putting $u = tan(x),$ and $tan(2x) = \frac{2u}{1-u^2}$ so that $-u^3 + u^2 + 3u = 1,$ but I can't find any roots that would help me.

I have also tried using all the trigonometric identities I could think of but that hasn't helped me either, so I have a feeling that I should "see" something that I am failing to see. Any advice?

Edit: I should find all real solutions.

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I think you may have just made an algebraic slip: $$\frac{2u}{1-u^2} = 1+u$$ implies $$2u = (1+u)(1-u^2)= 1 -u^3 - u^2 + u$$ and that $$u^3 + u^2 +u -1 = 0.$$

Can you continue from here?

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  • $\begingroup$ For self-check, one (and the only) solution is $(\sqrt[3]{17 + \sqrt{297}} + \sqrt[3]{17-\sqrt{297}} - 1) / 3$ $\endgroup$ – Abstraction Aug 9 '16 at 8:33
  • $\begingroup$ @Zestylemonzi I wish I could say that I can continue from there but I still can't seem to find a real solution to that equation that helps me... $\endgroup$ – t-brum Aug 9 '16 at 8:45
  • $\begingroup$ @t-brum From the form of the solution, it seems that you have to use something like Cardano's method (en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method), there are no short-cuts here. $\endgroup$ – Abstraction Aug 9 '16 at 9:17
  • $\begingroup$ Ok, I don't think that I am "supposed to" need such a method (this being the second chapter in my first university course and nothing about Cardano's method has been mentioned). :) Do you know if there is there any other way to go about solving this? @Abstraction $\endgroup$ – t-brum Aug 9 '16 at 9:23

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