since the given determinant is equal to zera

$\Rightarrow 0(0-\cos x \sin x)-\cos x\left(0-\cos ^{2} x\right)$

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

$\Rightarrow \cos ^{3} x-\sin ^{3} x=0$

$\Rightarrow \tan ^{3}=1 \Rightarrow \tan x=1$

$\therefore \quad \sum_{x \in s} \tan \left(\frac{\pi}{3}+x\right)=\sum_{x \in s} \frac{\tan \pi / 3+\tan x}{1-\tan \pi / 3 \cdot \tan x}$

${ = \sum\limits_{x\, \in \,s} {\frac{{\sqrt 3 \, + \,1}}{{1\, – \,\sqrt 3 }}\, = \sum\limits_{x\, \in \,s} {\frac{{\sqrt 3 \, + \,1}}{{1\, – \,\sqrt 3 }}\, \times \,\frac{{1 + \sqrt 3 }}{{1 + \sqrt 3 }}} \,} }$

${ \Rightarrow \sum\limits_{x \in s} {\frac{{1 + 3 + 2\sqrt 3 }}{{ – 2}}}  =  – 2 – \sqrt 3 }$