If $2{\tan }^2\theta -5\sec \theta =1$ has exactly 7 solutions in the interval $[\theta , n \mathrm{\pi }/ 2] , n \in N,$then the least and greatest values of n 

we have,

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

$\Rightarrow \sec \theta =3 [\because |\sec \theta |\ge 1\therefore 2\sec \theta +1\ne 0]$

 We observe that the curves $y=sec x$and $y=3$intersect at two points in $[0, 2 \mathrm{\pi }]$ one point lying in $(0, \mathrm{\pi }/2)$and the other in$( 3\mathrm{\pi }/2, 2\mathrm{\pi })$. Since $sec x$ is periodic with period $2 \mathrm{\pi }$. So the curves will intersect at two points in each of the intervals $[2\mathrm{\pi }, 4\mathrm{\pi }]$ and $[4\mathrm{\pi }, 6\mathrm{\pi }]$. Thus, $y =sec x$ and $y=3$intersect in 6 points in $[0, 6\mathrm{\pi }]$. Clearly, 7th point of intersection lies in the interval $( 6\mathrm{\pi }, 6\mathrm{\pi } + \frac{\mathrm{\pi }}{2})$or $\{6\mathrm{\pi },6\mathrm{\pi }+\frac{3\mathrm{\pi }}{2}\}$Thus, the two curves will intersect at 7 points in 

$[0, 13 \mathrm{\pi }/2]$ and also in $[0, 15 \mathrm{\pi }/2]$. Hence, the least value of n is 13 and the greatest value of $n$ is 15.