Find the values of `alpha` such that the variable point `(alpha, "tan" alpha)` lies inside the triangle whose sides are
`y=x+sqrt(3)-(pi)/(3), x+y+(1)/(sqrt(3))+(pi)/(6) = 0 " and " x-(pi)/(2) = 0`

Explanation

To find the values of \( \alpha \) such that the point \( (\alpha, \tan \alpha) \) lies inside the triangle formed by the lines:

1. \( y = x + \sqrt{3} - \frac{\pi}{3} \)
2. \( x + y + \frac{1}{\sqrt{3}} + \frac{\pi}{6} = 0 \)
3. \( x - \frac{\pi}{2} = 0 \)

we will follow these steps:

### Step 1: Rewrite the equations of the lines

1. **Line 1**: \( y = x + \sqrt{3} - \frac{\pi}{3} \)
2. **Line 2**: Rearranging \( x + y + \frac{1}{\sqrt{3}} + \frac{\pi}{6} = 0 \) gives:
\[ y = -x - \frac{1}{\sqrt{3}} - \frac{\pi}{6} \]
3. **Line 3**: This line is vertical at \( x = \frac{\pi}{2} \).

### Step 2: Determine the vertices of the triangle

To find the vertices of the triangle formed by these lines, we need to find the intersection points of the lines.

1. **Intersection of Line 1 and Line 2**:
\[ x + \sqrt{3} - \frac{\pi}{3} = -x - \frac{1}{\sqrt{3}} - \frac{\pi}{6} \]
Rearranging gives:
\[ 2x + \sqrt{3} + \frac{1}{\sqrt{3}} - \frac{\pi}{3} + \frac{\pi}{6} = 0 \]
Solving for \( x \):
\[ 2x = -\left(\sqrt{3} + \frac{1}{\sqrt{3}} - \frac{\pi}{6}\right) \]
\[ x = -\frac{1}{2} \left(\sqrt{3} + \frac{1}{\sqrt{3}} - \frac{\pi}{6}\right) \]
Substitute \( x \) back into either line to find \( y \).

2. **Intersection of Line 1 and Line 3**:
Set \( x = \frac{\pi}{2} \) in Line 1:
\[ y = \frac{\pi}{2} + \sqrt{3} - \frac{\pi}{3} \]

3. **Intersection of Line 2 and Line 3**:
Set \( x = \frac{\pi}{2} \) in Line 2:
\[ y = -\frac{\pi}{2} - \frac{1}{\sqrt{3}} - \frac{\pi}{6} \]

### Step 3: Determine the conditions for \( (\alpha, \tan \alpha) \)

The point \( (\alpha, \tan \alpha) \) must satisfy the following inequalities:

1. **Above Line 1**:
\[ \tan \alpha > \alpha + \sqrt{3} - \frac{\pi}{3} \]

2. **Below Line 2**:
\[ \tan \alpha < -\alpha - \frac{1}{\sqrt{3}} - \frac{\pi}{6} \]

3. **To the left of Line 3**:
\[ \alpha < \frac{\pi}{2} \]

### Step 4: Solve the inequalities

Now we need to solve these inequalities for \( \alpha \).

1. **For Line 1**:
Rearranging gives:
\[ \tan \alpha - \alpha > \sqrt{3} - \frac{\pi}{3} \]

2. **For Line 2**:
Rearranging gives:
\[ \tan \alpha + \alpha < -\frac{1}{\sqrt{3}} - \frac{\pi}{6} \]

### Step 5: Analyze the results

The values of \( \alpha \) that satisfy all three inequalities will be the required values.