If S = \ x ∈ [0, 2π]: | cases 0 & cos x & - sin x sin x & 0 & cos x...

Given,\left|\begin{cases}0 & cos x & - sin x \\ sin x & 0 & cos x \\ cos x & sin x & 0\end{cases}\right| = 0

Expanding the determinant,

$⇒ 0 ( 0 - cos xsin x ) - cos x\left(0 - \cos^{2} x\right) - sin x\left(\sin^{2} x - 0\right) = 0$

$⇒$ $\cos^{3} x - \sin^{3} x = 0$

$⇒$ $\tan^{3} x = 1 ⇒ tan x = 1 ⇒ x = \dfrac{\pi}{4}, \dfrac{5\pi}{4} ⇒ S = \left\{\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right\}$

$∴ \mathop{∑}_{x \in S}\tan \left(\dfrac{\pi}{3} + x\right)$

$= \mathop{∑}_{x \in S}\left(\dfrac{\sqrt{3} + \tan x}{1 - \sqrt{3}\tan x}\right)$

$= 2\left(\dfrac{\sqrt{3} + 1}{1 - \sqrt{3}}\right) = - 4 - 2\sqrt{3}$