Solve `tan^(3) x+3 gt 3tan x+tan^(2) x`.

Explanation

To solve the inequality \( \tan^3 x + 3 > 3 \tan x + \tan^2 x \), we will follow these steps:

### Step 1: Rearranging the Inequality
We start by rearranging the inequality to one side:

\[ \tan^3 x - \tan^2 x - 3 \tan x + 3 > 0 \]

### Step 2: Factoring the Expression
Next, we will factor the left-hand side. We can group the terms:

\[ (\tan^3 x - \tan^2 x) + (-3 \tan x + 3) > 0 \]

Now, we can factor out common terms:

\[ \tan^2 x (\tan x - 1) - 3(\tan x - 1) > 0 \]

This can be factored further:

\[ (\tan x - 1)(\tan^2 x - 3) > 0 \]

### Step 3: Finding Critical Points
Now we will find the critical points by setting each factor to zero:

1. \( \tan x - 1 = 0 \) gives us \( \tan x = 1 \) or \( x = \frac{\pi}{4} + n\pi \) where \( n \in \mathbb{Z} \).
2. \( \tan^2 x - 3 = 0 \) gives us \( \tan x = \sqrt{3} \) or \( \tan x = -\sqrt{3} \). This leads to:
- \( \tan x = \sqrt{3} \) gives \( x = \frac{\pi}{3} + n\pi \)
- \( \tan x = -\sqrt{3} \) gives \( x = -\frac{\pi}{3} + n\pi \)

### Step 4: Testing Intervals
Now we will test the intervals determined by the critical points:

1. \( (-\infty, -\frac{\pi}{3}) \)
2. \( (-\frac{\pi}{3}, \frac{\pi}{4}) \)
3. \( (\frac{\pi}{4}, \frac{\pi}{3}) \)
4. \( (\frac{\pi}{3}, \infty) \)

Choose test points in each interval to determine the sign of the product \( (\tan x - 1)(\tan^2 x - 3) \).

### Step 5: Conclusion
After testing the intervals, we find:

- The product is positive in the intervals:
- \( (-\frac{\pi}{3}, \frac{\pi}{4}) \)
- \( (\frac{\pi}{3}, \infty) \)

Thus, the solution to the inequality is:

\[ x \in \left(-\frac{\pi}{3} + n\pi, \frac{\pi}{4} + n\pi\right) \cup \left(\frac{\pi}{3} + n\pi, \frac{\pi}{2} + n\pi\right) \]

where \( n \in \mathbb{Z} \).