có 0<x<$\frac{\pi }{3}$ => sin(x)>0 và cos(x)>0có $\sin x=\frac{1}{\sqrt{3}}$ $\Rightarrow$ $\sin x^{2}=\frac{1}{3}$ $\Rightarrow$ $\cos x^{2}=\frac{2}{3}$ $\Rightarrow$ $\cos x=\sqrt{\frac{2}{3}}$
$\cos x+\frac{\pi }{6}=\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x$ = $\frac{\sqrt{3}}{2}.\sqrt{\frac{2}{3}}-\frac{1}{2}.\frac{1}{\sqrt{3}}$ = $\frac{-\sqrt{3}+3\sqrt{2}}{6}$