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

Options

A. `4+2sqrt(3)`
B. `-2+sqrt(3)`
C. `-2-sqrt(3)`
D. `-4-2sqrt(3)`

Answer

To solve the problem, we need to find the set \( S \) defined by the condition that the determinant of the given matrix is zero. The matrix is:

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

### Step 1: Calculate the Determinant

We can calculate the determinant using the formula for a \( 3 \times 3 \) matrix:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix, we have:

- \( a = 0 \)
- \( b = \cos x \)
- \( c = -\sin x \)
- \( d = \sin x \)
- \( e = 0 \)
- \( f = \cos x \)
- \( g = \cos x \)
- \( h = \sin x \)
- \( i = 0 \)

Substituting these values into the determinant formula:

\[ D = 0 \cdot (0 \cdot 0 - \cos x \cdot \sin x) - \cos x \cdot (\sin x \cdot 0 - \cos x \cdot \cos x) - \sin x \cdot (\sin x \cdot \sin x - 0 \cdot \cos x) \]

This simplifies to:

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

Thus,

\[ D = \cos^3 x - \sin^3 x \]

### Step 2: Set the Determinant to Zero

We need to find when the determinant is zero:

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

This can be factored using the difference of cubes:

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

Since \( \cos^2 x + \sin^2 x = 1 \), the second factor is always positive. Therefore, we only need to solve:

\[ \cos x - \sin x = 0 \]

### Step 3: Solve for \( x \)

This gives us:

\[ \cos x = \sin x \]

This occurs at:

\[ x = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z} \]

Within the interval \( [0, 2\pi] \), the solutions are:

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

Thus, the set \( S = \left\{ \frac{\pi}{4}, \frac{5\pi}{4} \right\} \).

### Step 4: Calculate the Summation

We need to find:

\[ \sum_{x \in S} \tan\left(\frac{\pi}{3} + x\right) \]

Calculating for each \( x \):

1. For \( x = \frac{\pi}{4} \):

\[ \tan\left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \tan\left(\frac{4\pi}{12} + \frac{3\pi}{12}\right) = \tan\left(\frac{7\pi}{12}\right) \]

2. For \( x = \frac{5\pi}{4} \):

\[ \tan\left(\frac{\pi}{3} + \frac{5\pi}{4}\right) = \tan\left(\frac{4\pi}{12} + \frac{15\pi}{12}\right) = \tan\left(\frac{19\pi}{12}\right) \]

### Step 5: Use the Tangent Addition Formula

Using the tangent addition formula:

\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]

Where \( a = \frac{\pi}{3} \) and \( b = x \).

1. For \( x = \frac{\pi}{4} \):

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

2. For \( x = \frac{5\pi}{4} \):

\[ \tan\left(\frac{19\pi}{12}\right) = \tan\left(\frac{7\pi}{12} + \pi\right) = \tan\left(\frac{7\pi}{12}\right) \]

### Step 6: Combine the Results

Thus, the total sum is:

\[ \tan\left(\frac{7\pi}{12}\right) + \tan\left(\frac{19\pi}{12}\right) = \tan\left(\frac{7\pi}{12}\right) - \tan\left(\frac{7\pi}{12}\right) = 0 \]

### Final Answer

The final result is:

\[ \sum_{x \in S} \tan\left(\frac{\pi}{3} + x\right) = 0 \]