Solving an equation having trignometric functions on RHS and an number in LHS

Question

If $$\cos\frac\pi{2n}+\sin\frac\pi{2n}=\frac{\sqrt n}2, n\in\Bbb N,$$ find $n$.

I haven't been able to get an answer to this problem but I know that since the range of $$\sin x+\cos x$$ is $1,√2$ So the value of $n$ must lie between $4,8$.Ofcourse I can just check for all values but is there a way to do it analytically?

Answer 1

This probably steers away from the intention of the problem, but you can square both sides of the equation and take it from there.

\begin{align} \cos\frac\pi{2n}+\sin\frac\pi{2n}&=\frac{\sqrt{n}}2\\ 1+\sin\frac\pi n&=\frac n4\\ \sin\frac\pi n&=\frac n4-1 \end{align}

For $n\in\Bbb N$, we have LHS in range $(0,1]$. This limits the possibilities of $n$ to $(4,8]$. On this interval, $\sin\frac\pi n-\frac n4+1$ is strictly decreasing; therefore, there can only be one solution. By inspection, $n=6$ is the only solution.

Answer 2

$$\dfrac{1}{\sqrt 2}\cos\dfrac{\pi}{n} + \dfrac{1}{\sqrt 2}\sin\dfrac{\pi}{n} = \dfrac{1}{\sqrt 2}\dfrac{\sqrt n}{2} $$

LHS is $$ \sin(\dfrac{\pi}{4}+\dfrac{\pi}{n})$$ with range $\pm 1$