The least difference between the roots of 

$4 \cos \mathrm{x}(2-3{\sin }^2\mathrm{x}) + (\cos 2\mathrm{x}+1)=0;0\le \mathrm{x}\le \frac{\mathrm{\pi }}{2}$

$$\begin{gathered} 4\cos \mathrm{x}(2-3 {\sin }^2\mathrm{x})+\cos 2\mathrm{x}+1=0 \\ 4\cos \mathrm{x}(2-3+3{\cos }^2\mathrm{x})+2{\cos }^2\mathrm{x}=0 \\ -4\cos \mathrm{x}+12{\cos }^3\mathrm{x}+2{\cos }^2\mathrm{x}=0 \\ 2\cos \mathrm{x}\left(6{\cos }^2\mathrm{x}+\cos \mathrm{x}-2\right)=0 \\ \cos \mathrm{x}=0 \left(2\cos \mathrm{x}-1\right) \left(3\cos \mathrm{x}+2\right)=0 \\ \cos \mathrm{x}=0, \cos \mathrm{x}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right., \cos \mathrm{x}=\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \\ \mathrm{x}=\frac{\mathrm{\pi }}{3} \end{gathered}$$