The number of all the possible integral values of n>2 such that...

Given, \sin $\frac{\pi}{2 n}+\cos \frac{\pi}{2 n}=\frac{\sqrt{n}}{2}$Squaring both sides \left(\sin \frac{\pi}{2 n}+\cos \frac{\pi}{2 n}\rright)^2=\frac{n}{4} \begin{gathered} \sin ^2 \frac{\pi}{2 n}+\cos ^2 \frac{\pi}{2 n}+2 \sin \frac{\pi}{2 n} \cos \frac{\pi}{2 n}=\frac{n}{4}\\ \Rightarrow 1+\sin \frac{\pi}{n}=\frac{n}{4} \Rightarrow \sin \frac{\pi}{n}=\frac{n}{4}-1 \Rightarrow \sin \frac{\pi}{n}=\frac{n-4}{4} \end{gathered}$As,$\quad \sin \frac{\pi}{n}>0, n>2$So,$\quad \frac{n-4}{4}>0 \Rightarrow n>4$Also,$\sin $\left(\frac{\pi}{2 n}\right)+\cos \frac{\pi}{2 n}=\sqrt{2} \sin \left(\frac{\pi}{4}+\frac{\pi}{2 n}\right). From Eq. (i) \sqrt{2} \sin \left(\frac{\pi}{4}+\frac{\pi}{2 n}\right)=\frac{\sqrt{n}}{2} \quad \sin \left(\frac{\pi}{4}+\frac{\pi}{2 n}\right)=\frac{\sqrt{n}}{2 \sqrt{2}} Since,$\sin \left(\frac{\pi}{4}+\frac{\pi}{2 n}\right)$< 1, \forall n>2$ We get, \frac{\sqrt{n}}{2 \sqrt{2}}$< 1 \Rightarrow n < 8$ From Eqs. (ii) and (iii) 4 < n < 8 So, $n=5,6,7$ Hence, number of integral value of $n$ is 3 .