Show that the equation \(\sin \left( x + \frac { 1 } { 6 } \pi \right) = \cos \left( x - \frac { 1 } { 4 } \pi \right)\) can be written in the form \(\tan x = \frac { \sqrt { 2 } - 1 } { \sqrt { 3 } - \sqrt { 2 } }\).
Hence solve the equation \(\sin \left( x + \frac { 1 } { 6 } \pi \right) = \cos \left( x - \frac { 1 } { 4 } \pi \right)\) for \(0 \leqslant x \leqslant 2 \pi\).