Solve ` tan x - tan^2 x gt 0and |2 sin x| lt 1` .

Answer

To solve the inequalities \( \tan x - \tan^2 x > 0 \) and \( |2 \sin x| < 1 \), we will break down the problem step by step.

### Step 1: Solve the first inequality \( \tan x - \tan^2 x > 0 \)

1. Rewrite the inequality:
\[ \tan x - \tan^2 x > 0 \]
Factor out \( \tan x \):
\[ \tan x (1 - \tan x) > 0 \]

2. Set \( y = \tan x \). The inequality becomes:
\[ y(1 - y) > 0 \]

3. Identify the critical points where the expression equals zero:
\[ y = 0 \quad \text{and} \quad y = 1 \]

4. Analyze the intervals determined by these critical points:
- The intervals are \( (-\infty, 0) \), \( (0, 1) \), and \( (1, \infty) \).

5. Test a point in each interval:
- For \( y < 0 \) (e.g., \( y = -1 \)): \( (-1)(1 - (-1)) = -1 \) (negative)
- For \( 0 < y < 1 \) (e.g., \( y = 0.5 \)): \( (0.5)(1 - 0.5) = 0.25 \) (positive)
- For \( y > 1 \) (e.g., \( y = 2 \)): \( (2)(1 - 2) = -2 \) (negative)

6. The solution for \( y(1 - y) > 0 \) is:
\[ 0 < y < 1 \quad \Rightarrow \quad 0 < \tan x < 1 \]

7. The values of \( x \) corresponding to this range are:
- \( \tan x = 0 \) at \( x = 0 \)
- \( \tan x = 1 \) at \( x = \frac{\pi}{4} \)

Thus, we have:
\[ 0 < x < \frac{\pi}{4} \]

### Step 2: Solve the second inequality \( |2 \sin x| < 1 \)

1. Rewrite the inequality:
\[ |2 \sin x| < 1 \]

2. This implies:
\[ -1 < 2 \sin x < 1 \]

3. Divide the entire inequality by 2:
\[ -\frac{1}{2} < \sin x < \frac{1}{2} \]

4. Identify the angles where \( \sin x = -\frac{1}{2} \) and \( \sin x = \frac{1}{2} \):
- \( \sin x = -\frac{1}{2} \) at \( x = -\frac{\pi}{6} + 2k\pi \) (for \( k \in \mathbb{Z} \))
- \( \sin x = \frac{1}{2} \) at \( x = \frac{\pi}{6} + 2k\pi \) (for \( k \in \mathbb{Z} \))

5. The solution for \( -\frac{1}{2} < \sin x < \frac{1}{2} \) in the interval \( [0, 2\pi] \) is:
\[ \frac{7\pi}{6} < x < \frac{11\pi}{6} \]

### Step 3: Find the intersection of the two solutions

1. The first solution gives:
\[ 0 < x < \frac{\pi}{4} \]

2. The second solution gives:
\[ \frac{7\pi}{6} < x < \frac{11\pi}{6} \]

3. The intersection of these two intervals is empty because \( \frac{7\pi}{6} \) is greater than \( \frac{\pi}{4} \).

### Final Solution

Thus, the solution set for the given inequalities is:
\[ \text{No solution} \]