#|(0,cosx,-sinx),(sinx,0,cosx),(cosx,sinx,0)|=0#.
Expanding by the first raw, we have,
#:. 0-cosx(0-cos^2x)-sinx(sin^2x-0)=0#.
# :. cos^3x-sin^3x=0, or, sin^3x=cos^3x......(star)#.
We must have #cos^3x!=0, because," otherwise "cosx=0#, and, then,
#(star) rArr sin^3x=cos^3x=0," so that, "sinx=0#.
# rArr cos^2x+sin^2x=0+0=0#, a contradiction.
So, dividing #(star)" by "cos^3x!=0, "we get, "tan^3x=1#.
#:. tanx=1, &, x in [0,2pi], x=pi/4,or, x=pi/4+pi=5/4pi#.
#:. S={pi/4,5pi/4} rArr AA x in S, tanx=1#.
#:. sum_( x in S)tan(pi/3+x)#,
#=tan(pi/3+pi/4)+tan(pi/3+5pi/4)#,
#={tan(pi/3)+tan(pi/4)}/{1-tan(pi/3)tan(pi/4)}#
#+{tan(pi/3)+tan(5pi/4)}/{1-tan(pi/3)tan(5pi/4)}#,
#=(sqrt3+1)/(1-sqrt3)+(sqrt3+1)/(1-sqrt3)#,
#={2(sqrt3+1)}/(1-sqrt3)xx(1+sqrt3)/(1+sqrt3)#,
#=2/{1-3)(1+sqrt3)^2#.
# rArr sum_( x in S)tan(pi/3+x)=-(1+2sqrt3+3)#,
showing that the right option is #(4): -4-2sqrt3#.
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