`2 "cos"(pi)/(2k) + 2 "cos"(pi)/(k) = sqrt3 + 1`
or `"cos"(pi)/(2k) + "cos"(pi)/(k) = (sqrt3 +1)/(2)`
Let `(pi)/(k) = theta`. Then,
`cos theta + "cos"(theta)/(2) = (sqrt3 +1)/(2)`
or `2t^(2) + t - (sqrt3 + 3)/(2) = 0` [where `cos (theta//2) = t`]
or `t = (-1 +- sqrt(1 + 4 (3 + sqrt3)))/(4)`
or `= (-1 +- (2 sqrt3 + 1))/(4) = (-2 -2 sqrt3)/(4), (sqrt3)/(2)`
`rArr t = "cos"(theta)/(2) in [-1, 1] :. "cos"(theta)/(2) = (sqrt3)/(2)`
`rArr (theta)/(2) = (pi)/(6) rArr k = 3`