If `s={x in [0,2pi]:|[0,cosx,-sinx],[sinx,0,cosx],[cosx,sinx,0]|=0}` then `sum_(x in s) tan(pi/3+x)` is equal to:

Options

A. `-2+sqrt3`
B. `4+2sqrt3`
C. `-4-2sqrt3`
D. `-2-sqrt3`

Answer

To solve the problem, we need to evaluate the determinant and find the values of \( x \) in the set \( s \) such that the determinant equals zero. Then, we will calculate the sum of \( \tan\left(\frac{\pi}{3} + x\right) \) for those values of \( x \).

### Step 1: Evaluate the Determinant
We are given the determinant:

\[ \begin{vmatrix} 0 & \cos x & -\sin x \\ \sin x & 0 & \cos x \\ \cos x & \sin x & 0 \end{vmatrix} \]

Calculating the determinant, we can use the cofactor expansion along the first row:

\[ D = 0 \cdot \begin{vmatrix} 0 & \cos x \\ \sin x & 0 \end{vmatrix} - \cos x \cdot \begin{vmatrix} \sin x & \cos x \\ \cos x & 0 \end{vmatrix} + (-\sin x) \cdot \begin{vmatrix} \sin x & 0 \\ \cos x & \sin x \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} \sin x & \cos x \\ \cos x & 0 \end{vmatrix} = -\sin^2 x \)
2. \( \begin{vmatrix} \sin x & 0 \\ \cos x & \sin x \end{vmatrix} = \sin^2 x \)

Thus, the determinant simplifies to:

\[ D = -\cos x (-\sin^2 x) + (-\sin x)(\sin^2 x) = \cos x \sin^2 x - \sin^3 x \]

Setting the determinant to zero:

\[ \cos x \sin^2 x - \sin^3 x = 0 \]

### Step 2: Factor the Equation
Factoring out \( \sin^2 x \):

\[ \sin^2 x (\cos x - \sin x) = 0 \]

This gives us two cases to consider:

1. \( \sin^2 x = 0 \)
2. \( \cos x - \sin x = 0 \)

### Step 3: Solve for \( x \)
**Case 1:** \( \sin^2 x = 0 \)

This implies \( \sin x = 0 \), which occurs at:

\[ x = 0, \pi, 2\pi \]

**Case 2:** \( \cos x - \sin x = 0 \)

This implies \( \cos x = \sin x \), which occurs at:

\[ x = \frac{\pi}{4}, \frac{5\pi}{4} \]

### Step 4: Combine Solutions
The complete set \( s \) is:

\[ s = \{ 0, \pi, 2\pi, \frac{\pi}{4}, \frac{5\pi}{4} \} \]

### Step 5: Calculate \( \sum_{x \in s} \tan\left(\frac{\pi}{3} + x\right) \)
We will evaluate \( \tan\left(\frac{\pi}{3} + x\right) \) for each \( x \) in \( s \).

1. For \( x = 0 \):
\[ \tan\left(\frac{\pi}{3} + 0\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]

2. For \( x = \pi \):
\[ \tan\left(\frac{\pi}{3} + \pi\right) = \tan\left(\frac{\pi}{3} + \pi\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]

3. For \( x = 2\pi \):
\[ \tan\left(\frac{\pi}{3} + 2\pi\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]

4. For \( x = \frac{\pi}{4} \):
\[ \tan\left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \tan\left(\frac{7\pi}{12}\right) \]
Using the tangent addition formula:
\[ \tan\left(\frac{7\pi}{12}\right) = \frac{\tan\left(\frac{\pi}{3}\right) + \tan\left(\frac{\pi}{4}\right)}{1 - \tan\left(\frac{\pi}{3}\right)\tan\left(\frac{\pi}{4}\right)} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \]

5. For \( x = \frac{5\pi}{4} \):
\[ \tan\left(\frac{\pi}{3} + \frac{5\pi}{4}\right) = \tan\left(\frac{7\pi}{12} + \pi\right) = \tan\left(\frac{7\pi}{12}\right) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \]

### Step 6: Sum the Values
Now we sum these values:

\[ \text{Total} = \sqrt{3} + \sqrt{3} + \sqrt{3} + \frac{\sqrt{3} + 1}{1 - \sqrt{3}} + \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \]

Calculating this gives:

\[ 3\sqrt{3} + 2 \cdot \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \]

### Final Answer
The final result is:

\[ \sum_{x \in s} \tan\left(\frac{\pi}{3} + x\right) = -4 - 2\sqrt{3} \]