The solution is not unique, but given that [tex]\sin x=\dfrac12[/tex], by the Pythagorean identity we have two solutions for [tex]\cos x[/tex]:
[tex]\sin^2x+\cos^2x=1\implies\cos x=\pm\sqrt{1-\sin^2x}=\pm\dfrac{\sqrt3}2[/tex]
Then from this, we have two corresponding solutions for [tex]\tan x[/tex]. By definition,
[tex]\tan x=\dfrac{\sin x}{\cos x}[/tex]
and so we can have, in the case of [tex]\cos x>0[/tex].
[tex]\tan x=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}[/tex]
or [tex]\tan x=-\dfrac1{\sqrt3}[/tex] in the case that [tex]\cos x<0[/tex].
