Number of solutions satisfying the equation $\frac{1}{\sin \mathrm{x}}-\frac{1}{\sin 2\mathrm{x}}=\frac{2}{\sin 4\mathrm{x}}\text{ in }[0,4\mathrm{\pi }]$ is 

$$\begin{gathered} \frac{1}{\sin \mathrm{x}}-\frac{1}{\sin {\displaystyle }2\mathrm{x}}=\frac{2}{\sin 4\mathrm{x}} \\ \begin{array}{l}\Rightarrow \frac{\sin 2\mathrm{x}-\sin \mathrm{x}}{\sin \mathrm{xsin} 2\mathrm{x}}=\frac{2}{2\sin 2\mathrm{xcos} 2\mathrm{x}}\\ \Rightarrow 2\sin 2\mathrm{xcos} 2\mathrm{x}-2\sin \mathrm{xcos} 2\mathrm{x}=2\sin \mathrm{x}\\ \Rightarrow \sin 4\mathrm{x}-\sin 3\mathrm{x}+\sin \mathrm{x}=2\sin \mathrm{x}\\ \Rightarrow \sin 4\mathrm{x}=\sin 3\mathrm{x}+\sin \mathrm{x}\\ \Rightarrow 2\sin 2\mathrm{xcos} 2\mathrm{x}=2\sin 2\mathrm{xcos} \mathrm{x} \text{ (as }\sin 2\mathrm{x}\ne 0\text{ ) }\\ \Rightarrow 2\mathrm{x}=2\mathrm{n\pi }\pm \mathrm{x}\end{array} \end{gathered}$$
$\begin{array}{l}\Rightarrow \mathrm{x}=\frac{2\mathrm{n\pi }}{3},\mathrm{n}\in \mathrm{I}\text{ (as }\mathrm{x}=2\mathrm{k\pi }\text{ is not in domain) }\\ \mathrm{n}=1,2,4,5,\dots .\end{array}$
Thus, four solution are $\frac{2\mathrm{\pi }}{3},\frac{4\mathrm{\pi }}{3},\frac{8\mathrm{\pi }}{3},\frac{10\mathrm{\pi }}{3}$.