Solve the general value.
`2 + sqrt3 sec x - 4 cos x = 2 sqrt3`

Answer

To solve the equation \( 2 + \sqrt{3} \sec x - 4 \cos x = 2 \sqrt{3} \), we will follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ 2 + \sqrt{3} \sec x - 4 \cos x = 2 \sqrt{3} \]

### Step 2: Multiply through by \( \cos x \)
To eliminate the secant function, multiply the entire equation by \( \cos x \):
\[ (2 + \sqrt{3} \sec x - 4 \cos x) \cos x = (2 \sqrt{3}) \cos x \]
This simplifies to:
\[ 2 \cos x + \sqrt{3} - 4 \cos^2 x = 2 \sqrt{3} \cos x \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ -4 \cos^2 x + 2 \cos x - 2 \sqrt{3} \cos x + \sqrt{3} = 0 \]
Combine like terms:
\[ -4 \cos^2 x + (2 - 2\sqrt{3}) \cos x + \sqrt{3} = 0 \]

### Step 4: Multiply by -1
To make the leading coefficient positive, multiply the entire equation by -1:
\[ 4 \cos^2 x - (2 - 2\sqrt{3}) \cos x - \sqrt{3} = 0 \]

### Step 5: Use the quadratic formula
This is a quadratic equation in the form \( ax^2 + bx + c = 0 \) where:
- \( a = 4 \)
- \( b = -(2 - 2\sqrt{3}) \)
- \( c = -\sqrt{3} \)

Using the quadratic formula \( \cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ b^2 = (2 - 2\sqrt{3})^2 = 4 - 8\sqrt{3} + 12 = 16 - 8\sqrt{3} \]
\[ 4ac = 4 \cdot 4 \cdot (-\sqrt{3}) = -16\sqrt{3} \]
Thus,
\[ b^2 - 4ac = (16 - 8\sqrt{3}) - (-16\sqrt{3}) = 16 - 8\sqrt{3} + 16\sqrt{3} = 16 + 8\sqrt{3} \]

Now substituting into the quadratic formula:
\[ \cos x = \frac{2 - 2\sqrt{3} \pm \sqrt{16 + 8\sqrt{3}}}{8} \]

### Step 6: Solve for \( \cos x \)
Calculate the values of \( \cos x \):
1. For \( \cos x = \frac{2 - 2\sqrt{3} + \sqrt{16 + 8\sqrt{3}}}{8} \)
2. For \( \cos x = \frac{2 - 2\sqrt{3} - \sqrt{16 + 8\sqrt{3}}}{8} \)

### Step 7: Find angles corresponding to \( \cos x \)
1. If \( \cos x = \frac{1}{2} \), then \( x = \frac{\pi}{3} + 2n\pi \) or \( x = -\frac{\pi}{3} + 2n\pi \).
2. If \( \cos x = -\frac{\sqrt{3}}{2} \), then \( x = \frac{5\pi}{6} + 2n\pi \) or \( x = -\frac{5\pi}{6} + 2n\pi \).

### Step 8: Write the general solution
The general solutions are:
\[ x = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z} \]
\[ x = 2n\pi \pm \frac{5\pi}{6}, \quad n \in \mathbb{Z} \]

### Final Answer
Thus, the general solution for the equation \( 2 + \sqrt{3} \sec x - 4 \cos x = 2 \sqrt{3} \) is:
\[ x = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z} \]
\[ x = 2n\pi \pm \frac{5\pi}{6}, \quad n \in \mathbb{Z} \]