The positive integer value of n>3 satisfying the equation 1sin⁡πn=1sin⁡2πn+1sin⁡3πn is

1sin⁡πn−1sin⁡3πn=1sin⁡2πnsin⁡3πn−sin⁡πnsin⁡πnsin⁡3πn=1sin⁡2πn2cos⁡2πnsin⁡πn=sin⁡πnsin⁡3πnsin⁡2πnsin⁡4πn=sin⁡3πn⇒4πn=π−3πn⇒n=7

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The positive integer value of $n > 3$ satisfying the equation $\frac{1}{\sin \frac{π}{n}} = \frac{1}{\sin \frac{2 π}{n}} + \frac{1}{\sin \frac{3 π}{n}}$ is

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answer is 7.

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Detailed Solution

$\begin{array}{l} \frac{1}{\sin \frac{π}{n}} - \frac{1}{\sin \frac{3 π}{n}} = \frac{1}{\sin \frac{2 π}{n}} \\ \frac{\sin \frac{3 π}{n} - \sin \frac{π}{n}}{\sin \frac{π}{n} \sin \frac{3 π}{n}} = \frac{1}{\sin \frac{2 π}{n}} \\ 2 \cos \frac{2 π}{n} \sin \frac{π}{n} = \frac{\sin \frac{π}{n} \sin \frac{3 π}{n}}{\sin \frac{2 π}{n}} \\ \sin \frac{4 π}{n} = \sin \frac{3 π}{n} \\ \Rightarrow \frac{4 π}{n} = π - \frac{3 π}{n} \end{array}$
$\Rightarrow$ n=7