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

Answer

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

### Step 1: Rearrange the Inequality
First, we rearrange the inequality to bring all terms to one side:
\[ \tan^3 x + 3 - 3 \tan x - \tan^2 x > 0 \]
This simplifies to:
\[ \tan^3 x - \tan^2 x - 3 \tan x + 3 > 0 \]

### Step 2: Substitute \( t = \tan x \)
Let \( t = \tan x \). The inequality now becomes:
\[ t^3 - t^2 - 3t + 3 > 0 \]

### Step 3: Factor the Polynomial
Next, we need to factor the polynomial \( t^3 - t^2 - 3t + 3 \). We can use the Rational Root Theorem to find possible rational roots. Testing \( t = 1 \):
\[ 1^3 - 1^2 - 3(1) + 3 = 1 - 1 - 3 + 3 = 0 \]
So, \( t = 1 \) is a root. We can factor \( t - 1 \) out of the polynomial.

Using synthetic division or polynomial long division, we divide \( t^3 - t^2 - 3t + 3 \) by \( t - 1 \):
\[ t^3 - t^2 - 3t + 3 = (t - 1)(t^2 + 0t - 3) \]
Thus, we have:
\[ t^3 - t^2 - 3t + 3 = (t - 1)(t^2 - 3) \]

### Step 4: Set the Factors Greater Than Zero
Now we need to solve:
\[ (t - 1)(t^2 - 3) > 0 \]
We can find the roots of \( t^2 - 3 = 0 \):
\[ t^2 = 3 \implies t = \sqrt{3} \quad \text{or} \quad t = -\sqrt{3} \]
Thus, the critical points are \( t = 1, \sqrt{3}, -\sqrt{3} \).

### Step 5: Test Intervals
We will test the intervals determined by these critical points: \( (-\infty, -\sqrt{3}), (-\sqrt{3}, 1), (1, \sqrt{3}), (\sqrt{3}, \infty) \).

1. **Interval \( (-\infty, -\sqrt{3}) \)**: Choose \( t = -2 \)
\[ (-2 - 1)(-2^2 - 3) = (-3)(4 - 3) = -3 < 0 \]
2. **Interval \( (-\sqrt{3}, 1) \)**: Choose \( t = 0 \)
\[ (0 - 1)(0^2 - 3) = (-1)(-3) = 3 > 0 \]
3. **Interval \( (1, \sqrt{3}) \)**: Choose \( t = 2 \)
\[ (2 - 1)(2^2 - 3) = (1)(4 - 3) = 1 > 0 \]
4. **Interval \( (\sqrt{3}, \infty) \)**: Choose \( t = 2 \)
\[ (2 - 1)(2^2 - 3) = (1)(4 - 3) = 1 > 0 \]

### Step 6: Combine the Results
The inequality is satisfied in the intervals:
\[ (-\sqrt{3}, 1) \cup (1, \sqrt{3}) \cup (\sqrt{3}, \infty) \]

### Step 7: Convert Back to \( x \)
Now we convert back to \( x \) using \( t = \tan x \):
- For \( t = -\sqrt{3} \): \( x = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \)
- For \( t = 1 \): \( x = \tan^{-1}(1) = \frac{\pi}{4} \)
- For \( t = \sqrt{3} \): \( x = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \)

Thus, the solution in terms of \( x \) is:
\[ x \in \left(-\frac{\pi}{3}, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{4}, \frac{\pi}{3}\right) \cup \left(\frac{\pi}{3}, \frac{\pi}{2}\right) \]

### Final Solution
The final solution is:
\[ x \in \left(-\frac{\pi}{3}, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{4}, \frac{\pi}{3}\right) \cup \left(\frac{\pi}{3}, \frac{\pi}{2}\right) \]