If $2 \tan ^ { 2 } \theta - 5 \sec \theta = 1$ has exactly $7$ solutions in the interval $\left[ 0 , \dfrac { n \pi } { 2 } \right] , n \in N$ then the least and greatest values of $n$ are

$\begin{array}{l}2 \tan ^{2} \theta-5 \sec \theta=1 \rightarrow 7 \text { solu.in }\left[0, \frac{n \pi}{2}\right] \\ \Rightarrow 2\left(\sec ^{2} \theta-1\right) 5 \sec \theta-1=0 \\ \Rightarrow 2 \sec ^{2} \theta-5 \sec \theta-3=0 \\ \sec \theta=\frac{5 \pm \sqrt{25+24}}{4} \\ =\frac{5 \pm 7}{4}=\frac{-1}{2}, 3 \\ \Rightarrow \sec \theta=3 \\ \Rightarrow \cos \theta=1 / 3 \rightarrow 7 \text { solu.in }\left[0, \frac{\mathrm{n} \pi}{2}\right] \\ \left(\frac{\mathrm{n} \pi}{2}\right)_{\min }=6 \pi+\pi / 2=\frac{13 \pi}{2} \\ \left(\frac{\mathrm{n} \pi}{2}\right)_{\max }=7 \pi+\pi / 2=\frac{15 \pi}{2} \\ \Rightarrow \mathrm{n}_{\text {min }}=13 \\ \mathrm{n}_{\max }=15\end{array}$