We need to examine three cases. 1) $CD=AD.$ This is the easiest of the cases because it immediately follows that $CD=BD$ and thus $x=\angle DCB=60^\circ$ and $\angle C=90^\circ.$ Therefore $\tan x=\sqrt{3}.$ 2) $CD=AC=b.$ We will need the law of sines and cosines in $\triangle BCD,$ the expression $4 CD^2=2(a^2+b^2)-c^2$ for the median, and the law of cosines in $\triangle DAC$ to find that $a=b\sqrt{7},$ $c=2b\sqrt{3},$ $\sin x=\frac{\sqrt{21}}{14},$ and $\cos x=\frac{5\sqrt{7}}{14}.$ Therefore $\tan x= \frac{\sqrt{3}}{5}.$ 3) $AD=AC.$ Similarly as in case #2 above, we get $\tan x=\frac{2+\sqrt{3}}{3}.$