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Find the general solution

Posted January 12th, 2010 by alpha
in

$ \tan\theta+\tan(\theta+\frac{\pi}{3})+\tan(\theta+\frac{2\pi}{3})=-3\tan\theta \tan(\theta+\frac{\pi}{3})\tan(\theta+\frac{2\pi}{3})=0 $

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Comments

answer

On January 13th, 2010 Structure says:

$ \tan\theta=t $

$$t+\frac{t+\sqrt 3}{1-t\sqrt 3}+\frac{t-\sqrt 3}{1+t\sqrt 3}=t+\frac{8t}{1-3t^2}=\frac{3t(3-t^2)}{1-3t^2}=-3t\frac{t+\sqrt 3}{1-t\sqrt 3}\frac{t-\sqrt 3}{1+t\sqrt 3}$$

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