The positive integer value of n 3 satisfyin | vedclass

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(C) Given equation: $\frac{1}{\sin(\frac{\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})} + \frac{1}{\sin(\frac{3\pi}{n})}$Rearranging the terms: $\frac{1}{

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$7$

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(C) Given equation: $\frac{1}{\sin(\frac{\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})} + \frac{1}{\sin(\frac{3\pi}{n})}$Rearranging the terms: $\frac{1}{\sin(\frac{\pi}{n})} - \frac{1}{\sin(\frac{3\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})}$Using the formula $\sin(A) - \sin(B) = 2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$,we get:$\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})}$$\frac{2\cos(\frac{2\pi}{n})\sin(\frac{\pi}{n})}{\sin(\frac{\pi}{n})\sin(\frac{3\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})}$$\frac{2\cos(\frac{2\pi}{n})}{\sin(\frac{3\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})}$$2\sin(\frac{2\pi}{n})\cos(\frac{2\pi}{n}) = \sin(\frac{3\pi}{n})$Using $\sin(2	heta) = 2\sin	heta\cos	heta$,we have:$\sin(\frac{4\pi}{n}) = \sin(\frac{3\pi}{n})$Since $\sin(A) = \sin(B)$ implies $A = \pi - B$ (for $A, B$ in the relevant range),we have:$\frac{4\pi}{n} = \pi - \frac{3\pi}{n}$$\frac{4\pi}{n} + \frac{3\pi}{n} = \pi$$\frac{7\pi}{n} = \pi$$n = 7$

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

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