The general solution of the equation `tan^2 x + 2 sqrt3 tan x = 1`is given by

Options

A. `x = pi/2`
B. `(n + 1/2) pi`
C. `(6n + 1) pi/12`
D. `(n pi)/12`

Answer

To solve the equation \( \tan^2 x + 2\sqrt{3} \tan x = 1 \), we will follow these steps:

### Step 1: Rearranging the Equation
First, we rearrange the equation to set it to zero:
\[ \tan^2 x + 2\sqrt{3} \tan x - 1 = 0 \]

### Step 2: Identifying Coefficients
We can identify the coefficients for the quadratic equation \( Ax^2 + Bx + C = 0 \):
- Let \( y = \tan x \)
- Here, \( A = 1 \), \( B = 2\sqrt{3} \), and \( C = -1 \).

### Step 3: Applying the Quadratic Formula
We will use the quadratic formula to find \( y \):
\[ y = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Substituting the values of \( A \), \( B \), and \( C \):
\[ y = \frac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]

### Step 4: Simplifying the Expression
Calculating the discriminant:
\[ (2\sqrt{3})^2 = 4 \cdot 3 = 12 \]
\[ -4 \cdot 1 \cdot (-1) = 4 \]
Thus, the discriminant becomes:
\[ B^2 - 4AC = 12 + 4 = 16 \]
Now substituting back into the quadratic formula:
\[ y = \frac{-2\sqrt{3} \pm \sqrt{16}}{2} \]
\[ y = \frac{-2\sqrt{3} \pm 4}{2} \]

### Step 5: Finding the Two Solutions
We can simplify this further:
1. For the positive root:
\[ y_1 = \frac{-2\sqrt{3} + 4}{2} = -\sqrt{3} + 2 \]
2. For the negative root:
\[ y_2 = \frac{-2\sqrt{3} - 4}{2} = -\sqrt{3} - 2 \]

### Step 6: Finding the Angles
Now we need to find \( x \) such that:
1. \( \tan x = -\sqrt{3} + 2 \)
2. \( \tan x = -\sqrt{3} - 2 \)

For \( \tan x = -\sqrt{3} + 2 \):
- This value corresponds to an angle \( x = \frac{\pi}{12} \) (or \( 15^\circ \)).

For \( \tan x = -\sqrt{3} - 2 \):
- This value corresponds to an angle \( x = \frac{7\pi}{12} \) (or \( 105^\circ \)).

### Step 7: Writing the General Solution
The general solutions for \( x \) can be expressed as:
\[ x = \frac{\pi}{12} + n\pi \quad \text{and} \quad x = \frac{7\pi}{12} + n\pi \quad (n \in \mathbb{Z}) \]
Combining these, we can write:
\[ x = \frac{\pi}{12} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Final Answer
Thus, the general solution of the equation \( \tan^2 x + 2\sqrt{3} \tan x = 1 \) is:
\[ x = \frac{\pi}{12} + n\pi \quad (n \in \mathbb{Z}) \]