If  $\tan \theta +\tan (60°+\theta )+\tan (120°+\theta )=3,$then  $\theta =$

$\tan \theta +\tan (60°+\theta )+\tan (120°+\theta )=3$

$=\frac{\tan \theta }{3}+\frac{\sqrt{3}+\tan \theta }{1-\sqrt{3}\tan \theta }+\frac{-\sqrt{3}+\tan \theta }{1+\sqrt{3}\tan \theta }=3$

$=\frac{\tan \theta (1-3{\tan }^2\theta )+(\sqrt{3}+\tan \theta )(1+\sqrt{3}\tan \theta )+(\tan \theta -\sqrt{3})(1-\sqrt{3}\tan \theta )}{1-3{\tan }^2\theta }=3$

$=\frac{\tan \theta -3{\tan }^3\theta +\sqrt{3}+3\tan \theta +\tan \theta +\sqrt{3}{\tan }^2\theta +\tan \theta -\sqrt{3}{\tan }^2\theta -\sqrt{3}+3\tan \theta }{1-3{\tan }^2\theta }$

$=\frac{9\tan \theta -3{\tan }^3\theta }{1-3{\tan }^2\theta }=\frac{3}{1}$

$=\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }=1$

$3\theta =n\pi +\frac{\pi }{4}n\in Z$

$\theta =\frac{n\pi }{3}+\frac{\pi }{12}$

$=\frac{4n\pi +\pi }{12}$

$=(4n+1)\frac{\pi }{12}$