JEE Trigonometry

This problem can be easily done using argand plane.

Assume a circle of unit radii centered at origin of argand plane.

We know that, roots of equation ${ x }^{ n }$ -1=0 will lie on that circle and form a regular polygon of n vertices.

Now, distance between two vertices of such polygon is given by,

$\left| V_{1}\quad Vr \right| =\left| 2\sin { \frac { (r-1)\Pi }{ n } } \right|$

So, $\left| V_{1}\quad V_{2} \right| +\left| V_{1}\quad V_{3} \right| +......+\left| V_{1}\quad V_{7} \right| =\left| 2\sin { \frac { \Pi }{ n } } \right| +\left| 2\sin { \frac { 2\Pi }{ n } } \right| +......+\left| 2\sin { \frac { 7\Pi }{ n } } \right|$

They form an AP in angle of sine which can be written as

$2(\sin { \frac { 7\Pi }{ 2n } } \sin { \frac { 3\Pi }{ n } } )\quad /\quad \sin { \frac { \Pi }{ 24 } }$

Now , plugging this value in original equation we get

$2(\sin { \frac { 7\Pi }{ 2n } } \sin { \frac { 3\Pi }{ n } } )\quad /\quad (\sin { \frac { \Pi }{ 24 } \csc { \frac { \Pi }{ n } } } )=2\sin { \frac { \Pi }{ n } } (\frac { 1+\cot { \frac { \Pi }{ 24 } } }{ 2 } )$

which can be simplified to

$\frac { \sin { \frac { \Pi }{ 24 } } +\cos { \frac { \Pi }{ 24 } } }{ \sin { \frac { \Pi }{ 24 } } } =\frac { \cos { \frac { \Pi }{ 2n } } +\sin { (\frac { 13\Pi }{ 2n } -\frac { \Pi }{ 2 } } ) }{ \sin { \frac { \Pi }{ 2n } } }$

Comparing angles on both sides of equation we get $\boxed{n=12}$

I`ve jumped several steps as the solution was getting too long but if you want to know anything put it in the comment section.

Consider polygon to be inscribed in a circle of radius R. Hence each edge subtends an angle 2*pi/n at centre. Then use cosine rule to find each V1V3..etc. If radius at their ends make an angle x then its length is 2Rsin(x/2) . Hence we get an sine AP. Now compare RHS AND LHS to get n=12