The number of solution(s) of the equation${\sin }^3\mathrm{xcos}\mathrm{x}+{\sin }^2{\mathrm{xcos}}^2\mathrm{x}+\sin {\mathrm{xcos}}^3\mathrm{x}=1,\text{ in }[0,2\mathrm{\pi }]$is_______

$\begin{array}{l}\sin \mathrm{xcos}\mathrm{x}\left[{\sin }^2\mathrm{x}+\sin \mathrm{xcos}\mathrm{x}+{\cos }^2\mathrm{x}\right]=1\\ \Rightarrow \sin \mathrm{xcos}\mathrm{x}+(\sin \mathrm{xcos}\mathrm{x}{)}^2=1\\ {\sin }^{2}2\mathrm{x}+2\sin 2\mathrm{x}-4=0\\ \Rightarrow \sin 2\mathrm{x}=\frac{-2\pm \sqrt{4+16}}{2}\end{array}$
$=-1\pm \sqrt{5}$,which is not possible.