Ta có

\(2\cos \alpha +\sqrt{2}\sin \alpha =2\Leftrightarrow \sqrt{2}\sin \alpha =2-2\cos \alpha \to 2{{\sin }^{2}}\alpha ={{\left( 2-2\cos \alpha  \right)}^{2}}\)

\(\begin{array}{l}
 \Leftrightarrow 2{\sin ^2}\alpha  = 4 - 8\cos \alpha  + 4{\cos ^2}\alpha  \Leftrightarrow 2\left( {1 - {{\cos }^2}\alpha } \right) = 4 - 8\cos \alpha  + 4{\cos ^2}\alpha \\
 \Leftrightarrow 6{\cos ^2}\alpha  - 8\cos \alpha  + 2 = 0 \Leftrightarrow \left[ \begin{array}{l}
\cos \alpha  = 1\\
\cos \alpha  = \frac{1}{3}
\end{array} \right..
\end{array}\)

\(\bullet \) \(\cos \alpha =1\): không thỏa mãn vì \({{0}^{0}}<\alpha <{{90}^{0}}.\)

\(\bullet \) \(\cos \alpha =\frac{1}{3}\Rightarrow \sin \alpha =\frac{2\sqrt{2}}{3}\xrightarrow{{}}\cot \alpha =\frac{\cos \alpha }{\sin \alpha }=\frac{\sqrt{2}}{4}.\)