Let ` 2 sin^(2)x + 3 sin x -2 gt 0 and x^(2)-x -2 lt 0` ( x is measured in radian ) . Then 'x' lies in the internal .

Options

A. `((pi)/(6),(5pi)/(6))`
B. `(-1,(5pi)/(6))`
C. `(-1,2)`
D. `((pi)/(6),2)`

Answer

To solve the inequalities \( 2 \sin^2 x + 3 \sin x - 2 > 0 \) and \( x^2 - x - 2 < 0 \), we will follow these steps:

### Step 1: Solve the first inequality \( 2 \sin^2 x + 3 \sin x - 2 > 0 \)

1. **Rewrite the inequality**:
\[ 2 \sin^2 x + 3 \sin x - 2 > 0 \]

2. **Let \( y = \sin x \)**. The inequality becomes:
\[ 2y^2 + 3y - 2 > 0 \]

3. **Factor the quadratic**:
To factor \( 2y^2 + 3y - 2 \), we look for two numbers that multiply to \( 2 \times -2 = -4 \) and add to \( 3 \). The numbers \( 4 \) and \( -1 \) work:
\[ 2y^2 + 4y - y - 2 > 0 \]
Grouping gives:
\[ 2y(y + 2) - 1(y + 2) > 0 \]
Factoring out \( (y + 2) \):
\[ (2y - 1)(y + 2) > 0 \]

4. **Find the critical points**:
Set each factor to zero:
\[ 2y - 1 = 0 \Rightarrow y = \frac{1}{2} \]
\[ y + 2 = 0 \Rightarrow y = -2 \quad (\text{not valid since } y = \sin x \text{ ranges from } -1 \text{ to } 1) \]

5. **Test intervals**:
The critical point is \( y = \frac{1}{2} \). We test intervals:
- For \( y < \frac{1}{2} \) (e.g., \( y = 0 \)): \( (2(0) - 1)(0 + 2) = -1 \cdot 2 < 0 \)
- For \( y > \frac{1}{2} \) (e.g., \( y = 1 \)): \( (2(1) - 1)(1 + 2) = 1 \cdot 3 > 0 \)

Thus, the solution for \( 2y^2 + 3y - 2 > 0 \) is:
\[ y > \frac{1}{2} \quad \Rightarrow \quad \sin x > \frac{1}{2} \]

6. **Find the angles**:
The angles where \( \sin x > \frac{1}{2} \) are:
\[ x \in \left( \frac{\pi}{6}, \frac{5\pi}{6} \right) \]

### Step 2: Solve the second inequality \( x^2 - x - 2 < 0 \)

1. **Factor the quadratic**:
\[ x^2 - x - 2 = (x + 1)(x - 2) < 0 \]

2. **Find the critical points**:
Set each factor to zero:
\[ x + 1 = 0 \Rightarrow x = -1 \]
\[ x - 2 = 0 \Rightarrow x = 2 \]

3. **Test intervals**:
The critical points divide the number line into intervals:
- For \( x < -1 \) (e.g., \( x = -2 \)): \( (-2 + 1)(-2 - 2) = (-1)(-4) > 0 \)
- For \( -1 < x < 2 \) (e.g., \( x = 0 \)): \( (0 + 1)(0 - 2) = (1)(-2) < 0 \)
- For \( x > 2 \) (e.g., \( x = 3 \)): \( (3 + 1)(3 - 2) = (4)(1) > 0 \)

Thus, the solution for \( x^2 - x - 2 < 0 \) is:
\[ -1 < x < 2 \]

### Step 3: Combine the solutions

1. **From the first inequality**:
\[ x \in \left( \frac{\pi}{6}, \frac{5\pi}{6} \right) \]

2. **From the second inequality**:
\[ x \in (-1, 2) \]

3. **Find the intersection**:
The intersection of \( \left( \frac{\pi}{6}, \frac{5\pi}{6} \right) \) and \( (-1, 2) \):
- Since \( \frac{\pi}{6} \approx 0.52 \) and \( \frac{5\pi}{6} \approx 2.62 \), the intersection is:
\[ x \in \left( \frac{\pi}{6}, 2 \right) \]

### Final Answer:
Thus, the solution for \( x \) lies in the interval:
\[ \boxed{\left( \frac{\pi}{6}, 2 \right)} \]