\(\sqrt3(\frac{1}{tanx}+tanx)\) = 4
\(\sqrt3(\frac{1+tan^2x}{tanx})\) = 4
√3+√3 tan2x = 4 tan x
√3 tan2x - 4 tan x+√3 = 0
Therefore
tan = √3 or tanx = \(\frac{1}{\sqrt3}\)
Therefore x = \(\frac{π}3\) or \(\frac{π}6\)
But here the smallest angle is \(\frac{π}6\)
Option C