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

We have    $2{\tan }^{2}x-5\sec x=1$

          $\begin{array}{l}\Rightarrow 2\left({\sec }^{2}x-1\right)-5\sec x=1\\ \Rightarrow 2{\sec }^{2}x-5\sec x-3=0\end{array}$

           $\Rightarrow \left(\sec x-3\right)\left(2\sec x+1\right)=0$

           $\Rightarrow \sec x=3 or secx=-\frac{1}{2}$

$\Rightarrow \cos x=\frac{1}{3}$

Which gives two values of x in each of  $\left[0,2\pi \right], (2\pi ,4\pi ], (4\pi ,6\pi ]$ and one value in $\left[6\pi ,6\pi +\frac{\pi }{2}\right]$ .

$\therefore$greatest value of n = 15.