1. Given equation

We need to solve $\sqrt{3}\sec x+\csc x+2(\tan x-\cot x)=0$ in $S=\{x\in(-\pi,\pi):x\neq 0,\pm \tfrac{\pi}{2}\}.$

Also, since $\sec x$ and $\csc x$ appear, we must have $\sin x\neq 0,\quad \cos x\neq 0,$ which is already ensured by the definition of $S$.

2. Rewrite in terms of $\sin x,\cos x$

Using $\sec x=\frac1{\cos x},\quad \csc x=\frac1{\sin x},\quad \tan x=\frac{\sin x}{\cos x},\quad \cot x=\frac{\cos x}{\sin x},$ we get $\frac{\sqrt3}{\cos x}+\frac1{\sin x}+2\left(\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}\right)=0.$

Multiply throughout by $\sin x\cos x$ (valid since neither is zero): $\sqrt3\sin x+\cos x+2(\sin^2x-\cos^2x)=0.$

Since $\sin^2x-\cos^2x=-\cos 2x,$ this becomes $\sqrt3\sin x+\cos x-2\cos 2x=0.$

3. Convert the linear trig part

Note that $\sqrt3\sin x+\cos x=2\sin\left(x+\frac\pi6\right),$ because $2\sin\left(x+\frac\pi6\right)=2\left(\sin x\cos\frac\pi6+\cos x\sin\frac\pi6\right)=\sqrt3\sin x+\cos x.$

So the equation becomes $2\sin\left(x+\frac\pi6\right)-2\cos 2x=0,$ so $\sin\left(x+\frac\pi6\right)=\cos 2x.$

Now use $\cos 2x=\sin\left(\frac\pi2-2x\right).$ Thus $\sin\left(x+\frac\pi6\right)=\sin\left(\frac\pi2-2x\right).$

4. Solve $\sin A=\sin B$

If $\sin A=\sin B$, then either $A=B+2n\pi$ or $A=\pi-B+2n\pi,$ where $n\in\mathbb Z$.

Here, $A=x+\frac\pi6,\qquad B=\frac\pi2-2x.$

Case 1:

$x+\frac\pi6=\frac\pi2-2x+2n\pi$ $3x=\frac\pi3+2n\pi$ $x=\frac\pi9+\frac{2n\pi}{3}.$

Case 2:

$x+\frac\pi6=\pi-\left(\frac\pi2-2x\right)+2n\pi$ $x+\frac\pi6=\frac\pi2+2x+2n\pi$ $-x=\frac\pi3+2n\pi$ $x=-\frac\pi3-2n\pi.$

Within $(-\pi,\pi)$, this gives only $x=-\frac\pi3.$

5. List all solutions in $(-\pi,\pi)$

From Case 1: $x=\frac\pi9+\frac{2n\pi}{3}.$ Check values in $(-\pi,\pi)$:

• $n=0$: $x=\frac\pi9$
• $n=1$: $x=\frac{7\pi}{9}$
• $n=-1$: $x=-\frac{5\pi}{9}$
• $n=2$: $x=\frac{13\pi}{9}>\pi$ not allowed
• $n=-2$: $x=-\frac{11\pi}{9}< -\pi$ not allowed

So Case 1 gives $x\in\left\{-\frac{5\pi}{9},\frac\pi9,\frac{7\pi}{9}\right\}.$

From Case 2: $x=-\frac\pi3.$

All are in $S$ and none equals $0,\pm \frac\pi2$.

Hence all distinct solutions are $\left\{-\frac{5\pi}{9},-\frac\pi3,\frac\pi9,\frac{7\pi}{9}\right\}.$

6. Find their sum

$-\frac{5\pi}{9}-\frac\pi3+\frac\pi9+\frac{7\pi}{9}$ Convert $\frac\pi3$ to ninths: $\frac\pi3=\frac{3\pi}{9}.$ So $\frac{-5\pi-3\pi+\pi+7\pi}{9}=\frac{0}{9}=0.$

Thus the required sum is $\boxed{0}.$

7. Option check

The correct option is $\boxed{\text{C: }0}.$