The number of solutions of the equation ${\sin }^5\mathrm{x}-{\cos }^5\mathrm{x}=\frac{1}{\cos \mathrm{x}}-\frac{1}{\sin \mathrm{x}}(\sin \mathrm{x}\ne \cos \mathrm{x})$ is 

Solution:

$\begin{array}{ll}\because & {\sin }^5\mathrm{x}-{\cos }^5\mathrm{x}=\frac{\sin \mathrm{x}-\cos \mathrm{x}}{\sin \mathrm{xcos}\mathrm{x}}\\ \Rightarrow & \sin \mathrm{xcos}\mathrm{x}\left[\frac{{\sin }^5\mathrm{x}-{\cos }^5\mathrm{x}}{\sin \mathrm{x}-\cos \mathrm{x}}\right]=1\\ \Rightarrow & \frac{1}{2}\sin 2\mathrm{x}\left[{\sin }^4\mathrm{x}+{\sin }^3\mathrm{xcos}\mathrm{x}+{\sin }^2{\mathrm{xcos}}^2\mathrm{x}\left.+\sin {\mathrm{xcos}}^2\mathrm{x}+{\cos }^4\mathrm{x}\right]=1\right.\\ & \begin{array}{ll}\Rightarrow & \sin 2\mathrm{x}\left[{\left({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}\right)}^2-2{\sin }^2{\mathrm{xcos}}^2\mathrm{x}\left.+\sin \mathrm{xcos}\mathrm{x}\left({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}\right)+{\sin }^2{\mathrm{xcos}}^2\mathrm{x}\right]=2\right.\end{array}\\ \Rightarrow & \sin 2\mathrm{x}\left[1-{\sin }^2{\mathrm{xcos}}^2\mathrm{x}+\sin \mathrm{xcos}\mathrm{x}\right]=2\\ \Rightarrow & {\sin }^{3}2\mathrm{x}-2{\sin }^{2}2\mathrm{x}-4\sin 2\mathrm{x}+8=0\\ \Rightarrow & (\sin 2\mathrm{x}-2{)}^2(\sin 2\mathrm{x}+2)=0\end{array}$

$\Rightarrow \sin 2\mathrm{x}=\pm 2,\text{ which is not possible for any }\mathrm{x}$