The positive integer value of n > 3 satisfying the equation 1/sin(π/n) = 1/sin(2π/n) + 1/sin(3π/n) is?

Given equation is
1/sin(π/n) = 1/sin(2π/n) + 1/sin(3π/n)
1/sin(π/n) - 1/sin(3π/n) = 1/sin(2π/n)
[sin(3π/n) - sin(π/n)] / [sin(3π/n)sin(π/n)] = 1/sin(2π/n)
[2cos(2π/n)sin(π/n)] / [sin(3π/n)sin(π/n)] = 1/sin(2π/n)
2cos(2π/n)sin(2π/n) = sin(3π/n)
sin(4π/n) = sin(3π/n)
The general solution for sinθ = sinα is given by
θ = pπ + (-1)^pα, p ∈ I
So, 4π/n = pπ + (-1)^p3π/n, p ∈ I
If p = 2m, then 4π/n = 2mπ + 3π/n
π/n = 2mπ or 1/n = 2m, which is not possible.
So, let p = 2m + 1 then 4π/n = (2m + 1)π - 3π/n
7π/n = (2m + 1)π
7/n = 2m + 1
For m = 0, n = 7
Hence, n = 7