\[\tan (x+\frac{\pi }{4})+\tan (x-\frac{\pi }{4})=\frac{\tan x+\tan \frac{\pi }{4}}{1-\tan x\tan \frac{\pi }{4}}+\frac{\tan x-\tan \frac{\pi }{4}}{1+\tan x\tan \frac{\pi }{4}}=\frac{\tan x+1}{1-\tan x}+\frac{\tan x-1}{1+\tan x}=\frac{{{(\tan x+1)}^{2}}+(\tan x-1)(1-\tan x)}{(1-\tan x)(1+\tan x)}=\frac{4\tan x}{1-{{\tan }^{2}}x}=2\sqrt{3}\]  با توجه به اتحاد مثلثاتی \[\tan 2x=\frac{2\tan x}{1-{{\tan }^{2}}x}\]، داریم: \[\tan 2x=\sqrt{3}=\tan \frac{\pi }{3}\Rightarrow 2x=k\pi +\frac{\pi }{3}\Rightarrow x=\frac{k\pi }{2}+\frac{\pi }{6}\]