The solution set of inequality
`((e^(x)-1)(2x-3)(x^(2)+x+2))/((sinx-2)(x+1)x) le 0`

Options

A. `[(3)/(2),oo)`
B. `(-oo,-1) uu [(3)/(2),oo)`
C. `(-1,0) uu [(3)/(2),oo)`
D. `R-{0,-1}`

Explanation

To solve the inequality

\[ \frac{(e^{x}-1)(2x-3)(x^{2}+x+2)}{(sinx-2)(x+1)x} \leq 0, \]

we will follow these steps:

### Step 1: Analyze the components of the inequality

1. **Identify the components**:
- Numerator: \( (e^{x}-1)(2x-3)(x^{2}+x+2) \)
- Denominator: \( (sinx-2)(x+1)x \)

### Step 2: Determine the sign of each component

2. **Numerator**:
- \( e^{x}-1 \): This is zero when \( x = 0 \) and positive for \( x > 0 \), negative for \( x < 0 \).
- \( 2x-3 \): This is zero when \( x = \frac{3}{2} \) and positive for \( x > \frac{3}{2} \), negative for \( x < \frac{3}{2} \).
- \( x^{2}+x+2 \): This is a quadratic with a positive leading coefficient and its discriminant \( 1^2 - 4(1)(2) = -7 \) is negative, meaning it is always positive.

3. **Denominator**:
- \( sinx-2 \): Since \( sinx \) ranges from -1 to 1, \( sinx-2 \) is always negative.
- \( x+1 \): This is zero when \( x = -1 \) and positive for \( x > -1 \), negative for \( x < -1 \).
- \( x \): This is zero when \( x = 0 \) and positive for \( x > 0 \), negative for \( x < 0 \).

### Step 3: Identify critical points

4. **Critical points**:
- From \( e^{x}-1 = 0 \): \( x = 0 \)
- From \( 2x-3 = 0 \): \( x = \frac{3}{2} \)
- From \( sinx-2 = 0 \): No real solutions
- From \( x+1 = 0 \): \( x = -1 \)
- From \( x = 0 \): \( x = 0 \)

### Step 4: Test intervals around the critical points

5. **Intervals**:
- We will test the intervals: \( (-\infty, -1) \), \( (-1, 0) \), \( (0, \frac{3}{2}) \), \( (\frac{3}{2}, \infty) \).

6. **Sign analysis**:
- For \( x < -1 \): All factors are negative, so the expression is positive.
- For \( -1 < x < 0 \): The numerator is negative and the denominator is negative, so the expression is positive.
- For \( 0 < x < \frac{3}{2} \): The numerator is negative and the denominator is negative, so the expression is positive.
- For \( x > \frac{3}{2} \): The numerator is positive and the denominator is negative, so the expression is negative.

### Step 5: Combine results

7. **Combine the intervals**:
- We want the expression to be less than or equal to zero. The intervals where this holds true are:
- \( (-\infty, -1) \) and \( (3/2, \infty) \).

### Step 6: Write the final solution

8. **Final solution**:
- The solution set of the inequality is:
\[ x \in (-\infty, -1) \cup \left[\frac{3}{2}, \infty\right). \]