If S = {x ∈ [0, 2π] : \begin{vmatrix} 0 & cosx & -sinx \ sinx & 0 & cosx \ cosx & sinx & 0 \end{vmatrix} = 0}, then Σ_{x∈S} tan(π/3 + x) is equal to

0(0-cosx) - cosx(0 - cos2x) - sinx(sin2x) = 0 ⇒ cos3x - sin3x = 0 ⇒ tan3x = 1 ⇒ tanx = 1
Σ√3 + tanx / 1 - √3tanx = Σ√3 + 1 / 1 - √3 * 1 + √3 / 1 + √3 = Σ1 + 3 + 2√3 / 2 = Σ2 + √3
This will happen twice. Once at π/4 and again at 5π/4 so the answer must be option C