If the sum of all the solutions of the equation $(2{\text{sec}}^{2}x-\text{cos}\frac{16}{9}x)\text{sec\hspace{0.17em}}x-(\text{cos}\frac{x}{3}+2\text{tan\hspace{0.17em}}x\cdot \text{sec\hspace{0.17em}}x)\text{tan\hspace{0.17em}}x=\text{sin}\frac{x}{3};x\in [-9\pi ,9\pi ]$   is  $k\pi$, then  $2k$ is equal to :

$2{\text{sec}}^{3}x-\text{cos}\frac{16}{9}x\cdot \text{sec}x-2{\text{tan}}^{2}x\cdot \text{sec}x=\text{sin}\frac{x}{3}+\text{cos}\frac{x}{3}\text{tan}x$

$$\begin{gathered} 2\text{sec}x-\text{cos}\frac{16}{9}x\cdot \text{sec}x=\frac{\text{sin}\frac{x}{3}\text{cos}x+\text{cos}\frac{x}{3}\text{sin}x}{\text{cos}x} \\ 2-\text{cos}\frac{16}{9}x=\text{sin}\frac{4x}{3} \\ \text{sin}\frac{4x}{3}+\text{cos}\frac{16x}{9}=2 \end{gathered}$$

$\Rightarrow \text{sin}\frac{4x}{3}=1 \text{cos}\frac{16x}{9}=1$

$\frac{4x}{3}=(4n+1)\frac{\pi }{2} \frac{16x}{9}=2m\pi$

$x=(4n+1)\frac{3\pi }{8} x=\frac{9m\pi }{8}$

$x=\frac{3\pi }{8},\frac{15\pi }{8},\frac{27\pi }{8},\dots x=0,\frac{9\pi }{8},\frac{18\pi }{8},\frac{27\pi }{8}$

Sum of common solution

$=\frac{27\pi }{8}+\frac{63\pi }{8}-\frac{9\pi }{8}-\frac{45\pi }{8}=\frac{36\pi }{8}\Rightarrow \text{k}\pi =\frac{36\pi }{8}\Rightarrow \text{k}=\frac{36}{8}=4.50$