If `A((sin alpha)/3 - 1,(cos alpha)/2 - 1)` and B(1,1) `alpha in [ -pi,pi]` are two points on the same side of the line `3x - 2y +1 = 0 ` then `alpha ` belongs to the interval

Options

A. `[-pi,-(3pi)/(4))uu((pi)/(4),pi]`
B. `[-pi,pi]`
C. `phi`
D. None of these

Answer

To solve the problem, we need to determine the interval for the angle \( \alpha \) such that the points \( A\left(\frac{\sin \alpha}{3} - 1, \frac{\cos \alpha}{2} - 1\right) \) and \( B(1, 1) \) are on the same side of the line given by the equation \( 3x - 2y + 1 = 0 \).

### Step-by-Step Solution:

1. **Identify the Line Equation**:
The line is given by:
\[ 3x - 2y + 1 = 0 \]
We can rewrite this as:
\[ 3x - 2y = -1 \]

2. **Substitute Point B into the Line Equation**:
We will first check the position of point \( B(1, 1) \) with respect to the line:
\[ 3(1) - 2(1) + 1 = 3 - 2 + 1 = 2 \]
Since \( 2 > 0 \), point \( B \) is above the line.

3. **Substitute Point A into the Line Equation**:
Now, we need to substitute point \( A \):
\[ A\left(\frac{\sin \alpha}{3} - 1, \frac{\cos \alpha}{2} - 1\right) \]
Substitute \( x = \frac{\sin \alpha}{3} - 1 \) and \( y = \frac{\cos \alpha}{2} - 1 \) into the line equation:
\[ 3\left(\frac{\sin \alpha}{3} - 1\right) - 2\left(\frac{\cos \alpha}{2} - 1\right) + 1 > 0 \]
Simplifying this:
\[ \sin \alpha - 3 - (\cos \alpha - 2) + 1 > 0 \]
\[ \sin \alpha - \cos \alpha > 0 \]

4. **Rearranging the Inequality**:
This can be rewritten as:
\[ \sin \alpha > \cos \alpha \]
Dividing both sides by \( \cos \alpha \) (assuming \( \cos \alpha \neq 0 \)):
\[ \tan \alpha > 1 \]

5. **Finding the Interval for \( \alpha \)**:
The inequality \( \tan \alpha > 1 \) implies:
\[ \alpha \in \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \]
However, since \( \alpha \) is restricted to the interval \( [-\pi, \pi] \), we need to adjust this interval:
\[ \alpha \in \left(\frac{\pi}{4}, \pi\right) \cup \left[-\pi, -\frac{3\pi}{4}\right) \]

6. **Final Interval**:
Thus, the values of \( \alpha \) for which points \( A \) and \( B \) are on the same side of the line are:
\[ \alpha \in \left[-\pi, -\frac{3\pi}{4}\right) \cup \left(\frac{\pi}{4}, \pi\right) \]