If  $2{\tan }^{2}x-5\sec x=1$ for exactly 10 distinct values of  $x\in \left[0,n\pi \right],n\in N$  then the greatest value of n is

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

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

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

Which gives two values of x in each of  $\left[0,2\pi \right],(2\pi ,4\pi ],(4\pi ,6\pi ],(6\pi ,8\pi ]$ and $twovaluesin(8\pi ,10\pi ]$

      $\therefore$   greatest value of n = 10.