Show that the equation
$$\tan \left( 45 ^ { \circ } + x \right) = 2 \tan \left( 45 ^ { \circ } - x \right)$$
can be written in the form
$$\tan ^ { 2 } x - 6 \tan x + 1 = 0$$
Hence solve the equation \(\tan \left( 45 ^ { \circ } + x \right) = 2 \tan \left( 45 ^ { \circ } - x \right)\), for \(0 ^ { \circ } < x < 90 ^ { \circ }\).