The maximum value of $\sin (x+\mathrm{\pi }/6)+\cos (\mathrm{x}+\mathrm{\pi }/6)$ in the interval of $(0,\mathrm{\pi }/2)$ is alttained at

$\sin (x+\mathrm{\pi }/6)+\cos (\mathrm{x}+\mathrm{\pi }/6)$ 

Mulitplying and dividing by $\sqrt{2}$

$$\begin{gathered} =\left(\sqrt{2}\right)\left[\right(1/\sqrt{2}\left)\sin \right(x+\pi /6)+(1/\sqrt{2}\left)\cos \right(x+\pi /6\left)\right] \\ =\sqrt{2}[\cos \pi /4\sin (x+\pi /6)+\sin \pi /4\cos (x+\pi /6\left)\right] \\ =\sqrt{2}\sin (x+\mathrm{\pi }/6+\mathrm{\pi }/4) \end{gathered}$$

Thus, $x=(\mathrm{\pi }/2)-(\mathrm{\pi }/4)=\mathrm{\pi }/12$