Solve the general value.
`tan ^(2) theta - (1 + sqrt3) tan theta + sqrt3 =0`

Answer

To solve the trigonometric equation \( \tan^2 \theta - (1 + \sqrt{3}) \tan \theta + \sqrt{3} = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \tan^2 \theta - (1 + \sqrt{3}) \tan \theta + \sqrt{3} = 0 \]

### Step 2: Let \( x = \tan \theta \)
We can substitute \( x \) for \( \tan \theta \) to simplify our equation:
\[ x^2 - (1 + \sqrt{3})x + \sqrt{3} = 0 \]

### Step 3: Use the quadratic formula
To solve for \( x \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = -(1 + \sqrt{3}) \), and \( c = \sqrt{3} \).

### Step 4: Calculate the discriminant
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = (1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3} \]
\[ 4ac = 4 \cdot 1 \cdot \sqrt{3} = 4\sqrt{3} \]
Thus, the discriminant is:
\[ D = (4 + 2\sqrt{3}) - 4\sqrt{3} = 4 - 2\sqrt{3} \]

### Step 5: Substitute back into the quadratic formula
Now we can substitute back into the quadratic formula:
\[ x = \frac{(1 + \sqrt{3}) \pm \sqrt{4 - 2\sqrt{3}}}{2} \]

### Step 6: Simplify the roots
Calculating \( \sqrt{4 - 2\sqrt{3}} \) can be complex, but we can directly find the roots by factoring or solving the quadratic equation.

### Step 7: Factor the equation
We can factor the quadratic as:
\[ (x - 1)(x - \sqrt{3}) = 0 \]
Thus, we have:
\[ x = 1 \quad \text{or} \quad x = \sqrt{3} \]

### Step 8: Solve for \( \theta \)
Now we solve for \( \theta \):
1. For \( x = 1 \):
\[ \tan \theta = 1 \implies \theta = 45^\circ + n \cdot 180^\circ \quad (n \in \mathbb{Z}) \]
In radians, this is:
\[ \theta = \frac{\pi}{4} + n\pi \]

2. For \( x = \sqrt{3} \):
\[ \tan \theta = \sqrt{3} \implies \theta = 60^\circ + n \cdot 180^\circ \quad (n \in \mathbb{Z}) \]
In radians, this is:
\[ \theta = \frac{\pi}{3} + n\pi \]

### Final Answer
The general solutions for \( \theta \) are:
\[ \theta = n\pi + \frac{\pi}{4} \quad \text{and} \quad \theta = n\pi + \frac{\pi}{3} \quad (n \in \mathbb{Z}) \]