Use the tangent addition formula: \( \tan(x+45^\circ)=\frac{\tan x+1}{\,1-\tan x\,}. \) Also \(\cot x=\dfrac{1}{\tan x}\).

Let \(t=\tan x\). Then \[ \frac{t+1}{1-t}=2\cdot\frac{1}{t} \;\Rightarrow\; t(t+1)=2(1-t) \] \[ \Rightarrow\; t^2+t=2-2t \;\Rightarrow\; t^2+3t-2=0. \]

Solve the quadratic: \[ t=\frac{-3\pm\sqrt{9+8}}{2} =\frac{-3\pm\sqrt{17}}{2} \approx 0.5616\;\text{ or }\;-3.5616. \]

\[ \tan x\approx0.5616 \Rightarrow x\approx \arctan(0.5616)=29.3^\circ, \] \[ \tan x\approx-3.5616 \Rightarrow x\text{ in QII } =180^\circ-\arctan(3.5616)=105.7^\circ. \]

Final answer

\(\boxed{x\approx29.3^\circ,\;105.7^\circ}\)