Question

The positive integer value of n>3 satisfying the equation

$\frac{1}{sin\left(\frac{\pi }{n}\right)}=\frac{1}{sin\left(\frac{2\pi }{n}\right)}+\frac{1}{sin\left(\frac{3\pi }{n}\right)}$ is ___

Answer 1

We have, $\frac{1}{sin\frac{\pi }{n}}-\frac{1}{sin\frac{3\pi }{n}}=\frac{1}{sin\frac{2\pi }{n}}\Rightarrow \frac{sin\frac{3\pi }{n}-sin\frac{\pi }{n}}{sin\frac{\pi }{n}sin\frac{3\pi }{n}}=\frac{1}{sin\frac{2\pi }{n}}\Rightarrow \frac{2cos\frac{2\pi }{n}sin\frac{\pi }{n}}{sin\frac{\pi }{n}sin\frac{3\pi }{n}}=\frac{1}{sin\frac{2\pi }{n}}\Rightarrow 2sin\frac{2\pi }{n}cos\frac{2\pi }{n}=sin\frac{3\pi }{n}\Rightarrow sin\frac{4\pi }{n}-sin\frac{3\pi }{n}=0\Rightarrow 2cos\frac{7\pi }{n}sin\frac{\pi }{n}=0\Rightarrow cos\frac{7\pi }{n}=0orsin\frac{\pi }{2n}=0\Rightarrow \frac{7\pi }{2n}=(2k+1)sin\frac{\pi }{2})or\frac{\pi }{2n}=2k\pi wherekϵZ\Rightarrow n=\frac{7}{2k+1}orn=\frac{1}{4k}$
($n=\frac{1}{4k}$ not possible for any integral value of k)
As n>3, for k=0, we get n=7