Trigonometry! #149

Square both sides of the equation (n is positive so both sides are positive) to get (using identities $\sin^2 x + \cos^2 x = 1$ and $2\sin x \cos x = \sin 2x$ )

$1+\sin(\frac {\pi}{n})=\frac {n}{4} \iff \sin(\frac {\pi}{n}) = \frac {n-4}{4}$

$\sin x \in [-1, 1]$ therefore $n \in (0, 8]$

Moreover left side is nonnegative because $\frac {\pi}{n} \in [0, \pi]$ therefore right side is also nonnegative so $n \in [4, 8]$ .

For values $n \in \{4, 5, 7, 8\}$ left side is irrational but left rational. For $n = 6$ the equation holds true thus this is the answer.

$\begin{aligned} \sin \left(\frac \pi {2n}\right) + \cos \left(\frac \pi {2n}\right) & = \frac {\sqrt n}2 & \small \color{#3D99F6} \text{Multiply both sides by }\frac 1{\sqrt 2}. \\ \frac 1{\sqrt 2} \sin \left(\frac \pi {2n}\right) + \frac 1{\sqrt 2} \cos \left(\frac \pi {2n}\right) & = \frac {\sqrt {2n}}4 \\ \sin \left(\frac \pi {2n} + \frac \pi 4\right) & = \frac {\sqrt{2n}}4 \end{aligned}$

For $n \ge 1$ , $\sin \left(\dfrac \pi {2n} + \dfrac \pi 4\right) > \sin \left(\dfrac \pi 4\right) > \dfrac 1{\sqrt 2}$ . Therefore, $\dfrac 1{\sqrt 2} < \dfrac {\sqrt{2n}}4 \le 1$ $\implies 4 < n \le 8$ . Only $n=\boxed{6}$ satisfies the equation $\sin \left(\dfrac \pi {2\color{#3D99F6}(6)} + \dfrac \pi 4\right) = \sin \left(\dfrac \pi 3 \right) = \dfrac {\sqrt 3}2 = \dfrac {\sqrt{2\color{#3D99F6}(6)}}4$ .

Let $x=\sin{\frac{\pi}{2n}}$ . Then $x+\sqrt{1-x^2}=\frac{\sqrt{n}}{2}\Rightarrow x^2=\frac{2\pm\sqrt{-\frac{n^2}{4}+2n}}{4}$ . The part under the square root has to be nonnegative, which leads to $n\leq 8$ . Then you could test all such $n$ to find $n=6$ . (No other trigonometric identities necessary from this point.)