If $2{\tan }^{2}x-5\sec x=1$  for exactly 7 distinct values of $x\in [0,\frac{n\pi }{2}], n\in N$ then the greatest value of n is 

$2{\tan }^{2}x-5\sec x=1$
$$\begin{gathered} \Rightarrow 2{\text{\hspace{0.17em}sec}}^{2}x-2-5\sec x-1=0 \\ \Rightarrow 2se{c}^{2}x-5secx-3=0 \\ \Rightarrow \left(2secx+1\right)(secx-3)=0 \end{gathered}$$
$\sec x=3\Rightarrow \cos x=\frac{1}{3}$

Which gives two values of  x  in each of $[0,2\pi ],(2\pi ,4\pi ],(4\pi ,6\pi ]$  and one value in 
 $6\pi +\frac{3\pi }{2}=15\frac{\pi }{2}$
$\therefore$ greatest value of  n = 15