The number of distinct solutions of the equation $\frac{5}{4}{\cos }^{2}2x+{\cos }^{4}x+{\sin }^{4}x+{\cos }^{6}x+{\sin }^{6}x=2$  in the interval $\left[0,2\pi \right]$  is

$$\begin{gathered} {\cos }^{4}x+{\sin }^{4}x=1-\frac{1}{2}.{\sin }^{2}2x \\ {\cos }^{6}x+{\sin }^{6}x=1-\frac{3}{2}{\sin }^{2}2x \\ \frac{5}{4}{\cos }^{2}2x+{\cos }^{4}x+{\sin }^{4}x+{\cos }^{6}x+{\sin }^{6}x=\frac{5}{4}{\cos }^{2}2x+2-2{\sin }^{2}2x \\ =\left(\frac{5}{4}+2\right){\cos }^{2}2x=\frac{13}{4}{\cos }^{2}2x \\ \Rightarrow \frac{13}{4}{\cos }^{2}2x=2\Rightarrow {\cos }^{2}2x=\frac{8}{13}\Rightarrow \cos 2x=\pm \sqrt{\frac{8}{13}} \\ Since x\in \left[0,2\mathrm{\pi }\right], \\ \Rightarrow we have 8 solutions. \end{gathered}$$