The number of solutions of ${\sin }^{2}x{\cos }^{2}x=1+{\cos }^{2}x{\sin }^{4}x$  in $[0,2\pi ]$ is

$We have {\sin }^{2}x{\cos }^{2}x={\cos }^{2}x{\sin }^{4}x+1$

$\Rightarrow {\sin }^{2}x{\cos }^{2}x-{\cos }^{2}x{\sin }^{4}x=1$

$\begin{array}{l}\Rightarrow {\sin }^{2}x{\cos }^{2}x-{\cos }^{2}x{\sin }^{2}x{\sin }^{2}x=1\\ \Rightarrow {\sin }^{2}x{\cos }^{2}x(1-{\sin }^{2}x)=1\\ \Rightarrow {\sin }^{2}x{\cos }^{4}x=1\end{array}$

No value of x can satisfy the above equation