The positive integer value of n 3 satisfying the equation 1 sin ( n ) = 1 sin ( 2 n ) + 1 sin ( 3 n ) is [2011]

Question

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 [2011]

Smart Achievers · Accepted Answer

(7)

$\frac{1}{\sin \frac{π}{n}} - \frac{1}{\sin \frac{3 π}{n}} = \frac{1}{\sin \frac{2 π}{n}}$

$\Rightarrow \frac{\sin \frac{3 π}{n} - \sin \frac{π}{n}}{\sin \frac{π}{n} \sin \frac{3 π}{n}} = \frac{1}{\sin \frac{2 π}{n}}$ $\Rightarrow \frac{2 \cos \frac{2 π}{n} \sin \frac{π}{n}}{\sin \frac{π}{n} \sin \frac{3 π}{n}} = \frac{1}{\sin \frac{2 π}{n}}$

$\Rightarrow 2 \sin \frac{2 π}{n} \cos \frac{2 π}{n} = \sin \frac{3 π}{n}$ $\Rightarrow \sin \frac{4 π}{n} - \sin \frac{3 π}{n} = 0$

$\Rightarrow 2 \cos \frac{7 π}{2 n} \sin \frac{π}{2 n} = 0 \Rightarrow \cos \frac{7 π}{2 n} = 0 or \sin \frac{π}{2 n} = 0$

$\Rightarrow \frac{7 π}{2 n} = (2 k + 1) \frac{π}{2} or \frac{π}{2 n} = 2 k π, where k \in ℤ$

$\Rightarrow n = \frac{7}{2 k + 1} or n = \frac{1}{4 k}$

$(n = \frac{1}{4 k} not possible for any integral value of k)$

$As n > 3; for k = 0, we get n = 7 .$

Solution

Q.
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 [2011]

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Solution

Q. The positive integer value of n>3 satisfying the equation 1sin(πn)=1sin(2πn)+1sin(3πn) is [2011]

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Q.
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 [2011]