Ответ:
Так как \(\cos 150^\circ=\cos(180^\circ-30^\circ)=-\cos 30^\circ=-\frac{\sqrt3}{2}\), получаем
\[\frac{\pi}{6}-\cos 150^\circ=\frac{\pi}{6}+\frac{\sqrt3}{2}=\frac{\pi+3\sqrt3}{6}.\]
Ответ: \(\frac{\pi+3\sqrt3}{6}\).
Так как \(\cos 150^\circ=\cos(180^\circ-30^\circ)=-\cos 30^\circ=-\frac{\sqrt3}{2}\), получаем
\[\frac{\pi}{6}-\cos 150^\circ=\frac{\pi}{6}+\frac{\sqrt3}{2}=\frac{\pi+3\sqrt3}{6}.\]
Ответ: \(\frac{\pi+3\sqrt3}{6}\).