Number of solutions of equation ${\sin }^2\mathrm{\theta }-\frac{4}{{\sin }^3\mathrm{\theta }-1}=1-\frac{4}{{\sin }^3\mathrm{\theta }-1},\mathrm{\theta }\in [0,6\mathrm{\pi }]$ is_________

$\begin{array}{l}\therefore {\sin }^2\mathrm{\theta }=1,\sin \mathrm{\theta }\ne 1\\ \Rightarrow \therefore \sin \mathrm{\theta }=-1\end{array}$
Thus, there are six values of $\mathrm{\theta }$ one each in intervals
$[0,2\mathrm{\pi }],[2\mathrm{\pi },4\mathrm{\pi }]\text{ and }[4\mathrm{\pi },6\mathrm{\pi }]$