$\begin{array}{l}
2{\sin ^2}x - 5\cos x + 5 = 0 \Leftrightarrow 2\left( {1 - {{\cos }^2}x} \right) - 5\cos x + 5 = 0\\
\Leftrightarrow - 2{\cos ^2}x - 5\cos x + 7 = 0 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = -\frac{7}{2}\left( {VN} \right)
\end{array} \right.\\
\Leftrightarrow x = k2\pi
\end{array}$