This problem requires a strong understanding of determinants, trigonometric equations, and trigonometric identities. Our strategy will be to first evaluate the given determinant and solve the resulting trigonometric equation to find the set of values for $x$. Then, we will use the tangent addition formula to calculate the sum of the required expressions.

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1. Understanding the Problem and Key Concepts

The problem asks us to:

• Find the set $S$ of all $x \in [0, 2\pi]$ for which the given $3 \times 3$ determinant is equal to zero.
• Calculate the sum of $\tan\left(\frac{\pi}{3} + x\right)$ for all $x$ in the set $S$.

We will use the following key concepts and formulas:

• Determinant of a $3 \times 3$ Matrix: For a matrix $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, its determinant is $a(ei - fh) - b(di - fg) + c(dh - eg)$.
• Trigonometric Identity: $\tan x = \frac{\sin x}{\cos x}$.
• Tangent Addition Formula: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
• Periodicity of Tangent Function: $\tan(\theta + n\pi) = \tan \theta$ for any integer $n$.

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2. Calculating the Determinant

We are given the determinant: \Delta = \left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right|

Let's expand this determinant along the first row: \Delta = 0 \cdot \left| {\matrix{ 0 & {\cos x} \cr {\sin x} & 0 \cr } } \right| - \cos x \cdot \left| {\matrix{ {\sin x} & {\cos x} \cr {\cos x} & 0 \cr } } \right| + (-\sin x) \cdot \left| {\matrix{ {\sin x} & 0 \cr {\cos x} & {\sin x} \cr } } \right|

• The first term is $0 \cdot (0 \cdot 0 - \cos x \cdot \sin x) = 0$.
• The second term is $-\cos x \cdot (\sin x \cdot 0 - \cos x \cdot \cos x) = -\cos x \cdot (0 - \cos^2 x) = -\cos x \cdot (-\cos^2 x) = \cos^3 x$.
• The third term is $-\sin x \cdot (\sin x \cdot \sin x - 0 \cdot \cos x) = -\sin x \cdot (\sin^2 x - 0) = -\sin x \cdot (\sin^2 x) = -\sin^3 x$.

Combining these terms, the determinant simplifies to: $\Delta = \cos^3 x - \sin^3 x$

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3. Solving the Trigonometric Equation

Now we set the determinant equal to zero to find the values of $x$: $\cos^3 x - \sin^3 x = 0$ $\cos^3 x = \sin^3 x$

Step-by-step reasoning:

• Why $\cos^3 x = \sin^3 x$: We isolate the trigonometric terms to prepare for solving.
• Why we can divide by $\cos^3 x$: If $\cos x = 0$, then $\sin x = \pm 1$ (since $\sin^2 x + \cos^2 x = 1$). In this case, $\cos^3 x = 0$ but $\sin^3 x = \pm 1$, so $\cos^3 x \ne \sin^3 x$. Therefore, $\cos x \ne 0$, which means we can safely divide both sides by $\cos^3 x$.

Dividing by $\cos^3 x$: $1 = \frac{\sin^3 x}{\cos^3 x}$ $1 = \left( {\frac{{\sin x}}{{\cos x}}} \right)^3$ $1 = \tan^3 x$

Taking the cube root of both sides: $\tan x = 1$

Now we need to find the values of $x$ in the interval $[0, 2\pi]$ for which $\tan x = 1$. The general solution for $\tan x = 1$ is $x = n\pi + \frac{\pi}{4}$, where $n$ is an integer.

• For $n=0$: $x = 0 \cdot \pi + \frac{\pi}{4} = \frac{\pi}{4}$. This is in $[0, 2\pi]$.
• For $n=1$: $x = 1 \cdot \pi + \frac{\pi}{4} = \frac{5\pi}{4}$. This is in $[0, 2\pi]$.
• For $n=2$: $x = 2 \cdot \pi + \frac{\pi}{4} = \frac{9\pi}{4}$. This is outside $[0, 2\pi]$.

So, the set $S$ is $\left\{ \frac{\pi}{4}, \frac{5\pi}{4} \right\}$.

Tip: Always remember to check the given interval for $x$ and find all solutions within that range. The periodicity of trigonometric functions is crucial here.

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4. Evaluating the Sum

We need to calculate $\sum\limits_{x \in S} {\tan \left( {{\pi \over 3} + x} \right)}$, which means we need to evaluate $\tan\left(\frac{\pi}{3} + x\right)$ for each $x \in S$ and sum the results.

We know $\tan(\pi/3) = \sqrt{3}$.

For $x = \frac{\pi}{4}$: Using the tangent addition formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$: $\tan\left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \frac{\tan(\pi/3) + \tan(\pi/4)}{1 - \tan(\pi/3)\tan(\pi/4)}$ Since $\tan(\pi/4) = 1$: $\tan\left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} = \frac{1 + \sqrt{3}}{1 - \sqrt{3}}$ To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, $(1+\sqrt{3})$: $\frac{1 + \sqrt{3}}{1 - \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}} = \frac{(1 + \sqrt{3})^2}{1^2 - (\sqrt{3})^2} = \frac{1 + 2\sqrt{3} + 3}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2}$ $= - (2 + \sqrt{3}) = -2 - \sqrt{3}$

For $x = \frac{5\pi}{4}$: We need to calculate $\tan\left(\frac{\pi}{3} + \frac{5\pi}{4}\right)$. Notice that $\frac{5\pi}{4} = \pi + \frac{\pi}{4}$. So, $\tan\left(\frac{\pi}{3} + \frac{5\pi}{4}\right) = \tan\left(\frac{\pi}{3} + \left(\pi + \frac{\pi}{4}\right)\right) = \tan\left(\left(\frac{\pi}{3} + \frac{\pi}{4}\right) + \pi\right)$. Since the tangent function has a period of $\pi$, $\tan(\theta + \pi) = \tan(\theta)$. Therefore: $\tan\left(\frac{\pi}{3} + \frac{5\pi}{4}\right) = \tan\left(\frac{\pi}{3} + \frac{\pi}{4}\right)$ This means the value is the same as the first term: $\tan\left(\frac{\pi}{3} + \frac{5\pi}{4}\right) = -2 - \sqrt{3}$

Summing the terms: The sum is the sum of these two values: $\sum\limits_{x \in S} {\tan \left( {{\pi \over 3} + x} \right)} = \left(-2 - \sqrt{3}\right) + \left(-2 - \sqrt{3}\right)$ $= -4 - 2\sqrt{3}$

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5. Final Answer and Summary

The calculated sum is $-4 - 2\sqrt{3}$. This corresponds to option (D).

Summary of Steps:

1. We expanded the $3 \times 3$ determinant and simplified it to $\cos^3 x - \sin^3 x$.
2. We set the determinant to zero, leading to the trigonometric equation $\tan^3 x = 1$, which simplifies to $\tan x = 1$.
3. We found the solutions for $\tan x = 1$ in the interval $[0, 2\pi]$ to be $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. This formed the set $S$.
4. For each $x \in S$, we calculated $\tan\left(\frac{\pi}{3} + x\right)$ using the tangent addition formula and rationalized the result. Both terms evaluated to $-2 - \sqrt{3}$.
5. Finally, we summed these values to get the total sum of $-4 - 2\sqrt{3}$.

The final answer is $\boxed{-\,\,4 - 2\sqrt 3}$.