The maximum value of $f(x)=\sin (x+\frac{\pi }{6})+\cos (x+\frac{\pi }{6})$ in the interval $[0,\frac{\pi }{2}]$ occurs at $x=$:

$f(x)=\sin (x+\frac{\pi }{6})+\cos (x+\frac{\pi }{6})$.We can rewrite this using $A\sin \theta +B\cos \theta =\sqrt{A^2+B^2}\sin (\theta +\varphi )$.$f(x)=\sqrt{1^2+1^2}\sin (x+\frac{\pi }{6}+\frac{\pi }{4})=\sqrt{2}\sin (x+\frac{5\pi }{12})$.The maximum value of the sine function occurs when the angle is $\frac{\pi }{2}$:$x+\frac{5\pi }{12}=\frac{\pi }{2}$$x=\frac{\pi }{2}-\frac{5\pi }{12}=\frac{6\pi -5\pi }{12}=\frac{\pi }{12}$.Since $\frac{\pi }{12}\in [0,\frac{\pi }{2}]$, the maximum occurs at $x=\frac{\pi }{12}$.