If $k\in I$  then the number of values of $x$  in $[0,2\pi ]$  which satisfy the equation   $\left(\frac{k}{2k-3}\right){\sin }^{2}x+\left(\frac{2k-3}{k}\right)\cos e{c}^{2}x=2$is ______

$$\begin{gathered} \underset{t}{\underbrace{\left(\frac{k}{2k-3}\right){\sin }^{2}x}}+\underset{\frac{1}{t}}{\underbrace{\left(\frac{2k-3}{k}\right)\cos e{c}^{2}x}}=2 \\ t+\frac{1}{t}=2\Rightarrow t=1 \\ \therefore \frac{k}{2k-3}{\sin }^{2}x=1\Rightarrow {\sin }^{2}x=\frac{2k-3}{k} \\ 0<\frac{2k-3}{k}\le 1\Rightarrow k\in [\frac{3}{2},3] \end{gathered}$$

  integral values of   are 2 & 3
For  $k=2$
 $2{\sin }^{2}x=1\Rightarrow \sin x=\pm \frac{1}{\sqrt{2}}\Rightarrow x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$
For  $k=3$
 ${\sin }^{2}x=1\Rightarrow \sin x=\pm 1\Rightarrow x=\frac{\pi }{2},\frac{3\pi }{2}$
  number of values of  $x$ in $[0,2\pi ]$  are 6