If the vectors `vec a` and `vec b`are linearly independent and satisfying `(sqrt3tantheta+1)vec a + (sqrt3sectheta-2)vec b=vec 0`,then the most general values of `theta` are:

Options

A. `npi - (pi)/(6), n in Z`
B. `2n pi pm (11pi)/(6), n in Z`
C. `n pi pm (pi)/(6), n in Z`
D. `2 npi + (11pi)/(6), n in Z`

Explanation

To solve the problem, we start with the given equation involving the vectors \(\vec{a}\) and \(\vec{b}\):

\[ (\sqrt{3} \tan \theta + 1) \vec{a} + (\sqrt{3} \sec \theta - 2) \vec{b} = \vec{0} \]

Since \(\vec{a}\) and \(\vec{b}\) are linearly independent, the only way for this linear combination to equal the zero vector is if both coefficients are zero. Therefore, we can set up the following equations:

1. \(\sqrt{3} \tan \theta + 1 = 0\)
2. \(\sqrt{3} \sec \theta - 2 = 0\)

### Step 1: Solve the first equation

From the first equation:

\[ \sqrt{3} \tan \theta + 1 = 0 \]

Rearranging gives:

\[ \sqrt{3} \tan \theta = -1 \]

Thus:

\[ \tan \theta = -\frac{1}{\sqrt{3}} \]

### Step 2: Find the angles for \(\tan \theta = -\frac{1}{\sqrt{3}}\)

The tangent function is negative in the second and fourth quadrants. The reference angle where \(\tan \theta = \frac{1}{\sqrt{3}}\) is \(\frac{\pi}{6}\). Therefore, the angles where \(\tan \theta = -\frac{1}{\sqrt{3}}\) are:

\[ \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \quad \text{(second quadrant)} \]
\[ \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \quad \text{(fourth quadrant)} \]

### Step 3: Solve the second equation

Now, we solve the second equation:

\[ \sqrt{3} \sec \theta - 2 = 0 \]

Rearranging gives:

\[ \sqrt{3} \sec \theta = 2 \]

Thus:

\[ \sec \theta = \frac{2}{\sqrt{3}} \]

### Step 4: Find the angles for \(\sec \theta = \frac{2}{\sqrt{3}}\)

The secant function is positive in the first and fourth quadrants. The corresponding cosine value is:

\[ \cos \theta = \frac{\sqrt{3}}{2} \]

The angles where \(\cos \theta = \frac{\sqrt{3}}{2}\) are:

\[ \theta = \frac{\pi}{6} \quad \text{(first quadrant)} \]
\[ \theta = \frac{11\pi}{6} \quad \text{(fourth quadrant)} \]

### Step 5: Combine the results

Now we have the angles from both equations:

From \(\tan \theta = -\frac{1}{\sqrt{3}}\):
- \(\theta = \frac{5\pi}{6} + n\pi\) for \(n \in \mathbb{Z}\)

From \(\sec \theta = \frac{2}{\sqrt{3}}\):
- \(\theta = \frac{\pi}{6} + 2k\pi\) or \(\theta = \frac{11\pi}{6} + 2k\pi\) for \(k \in \mathbb{Z}\)

The most general solutions for \(\theta\) that satisfy both conditions are:

\[ \theta = \frac{5\pi}{6} + n\pi \quad \text{and} \quad \theta = \frac{11\pi}{6} + 2k\pi \]

### Final Answer

The most general values of \(\theta\) are:

\[ \theta = \frac{5\pi}{6} + n\pi \quad \text{and} \quad \theta = \frac{11\pi}{6} + 2k\pi \]