If 1 lies between the roots of the equation `3x^(2)-3 sin alpha x -2 cos^(2) alpha =0`, the `alpha` lies in the interval

Options

A. `(0,(pi)/(2))`
B. `((pi)/(12),(pi)/(2))`
C. `((pi)/(6),(5 pi)/(6))`
D. `((pi)/(6), (pi)/(2)) cup ((pi)/(2),(5 pi)/(6))`

Answer

To solve the problem, we need to determine the interval of \( \alpha \) such that \( 1 \) lies between the roots of the quadratic equation \( 3x^2 - 3 \sin \alpha x - 2 \cos^2 \alpha = 0 \).

### Step 1: Identify the roots of the quadratic equation
The roots \( p \) and \( q \) of the quadratic equation \( ax^2 + bx + c = 0 \) can be expressed using the quadratic formula:
\[ p, q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, \( a = 3 \), \( b = -3 \sin \alpha \), and \( c = -2 \cos^2 \alpha \).

### Step 2: Use the condition that \( 1 \) lies between the roots
For \( 1 \) to lie between the roots \( p \) and \( q \), we need:
\[ f(1) < 0 \]
where \( f(x) = 3x^2 - 3 \sin \alpha x - 2 \cos^2 \alpha \).

### Step 3: Calculate \( f(1) \)
Substituting \( x = 1 \) into the function:
\[ f(1) = 3(1)^2 - 3 \sin \alpha (1) - 2 \cos^2 \alpha \]
\[ f(1) = 3 - 3 \sin \alpha - 2 \cos^2 \alpha \]

### Step 4: Set up the inequality
We need to solve:
\[ 3 - 3 \sin \alpha - 2 \cos^2 \alpha < 0 \]
Using the identity \( \cos^2 \alpha = 1 - \sin^2 \alpha \):
\[ 3 - 3 \sin \alpha - 2(1 - \sin^2 \alpha) < 0 \]
\[ 3 - 3 \sin \alpha - 2 + 2 \sin^2 \alpha < 0 \]
\[ 2 \sin^2 \alpha - 3 \sin \alpha + 1 < 0 \]

### Step 5: Factor the quadratic inequality
Factoring the quadratic:
\[ (2 \sin \alpha - 1)(\sin \alpha - 1) < 0 \]

### Step 6: Determine the critical points
The critical points are:
\[ \sin \alpha = \frac{1}{2} \quad \text{and} \quad \sin \alpha = 1 \]

### Step 7: Analyze the intervals
We need to test the intervals determined by these critical points:
1. \( \sin \alpha < \frac{1}{2} \)
2. \( \frac{1}{2} < \sin \alpha < 1 \)
3. \( \sin \alpha > 1 \)

### Step 8: Determine where the inequality holds
- For \( \sin \alpha < \frac{1}{2} \): Choose \( \sin \alpha = 0 \) (negative).
- For \( \frac{1}{2} < \sin \alpha < 1 \): Choose \( \sin \alpha = 0.6 \) (positive).
- For \( \sin \alpha > 1 \): Not possible since \( \sin \alpha \) cannot exceed 1.

Thus, the inequality \( (2 \sin \alpha - 1)(\sin \alpha - 1) < 0 \) holds for:
\[ \frac{1}{2} < \sin \alpha < 1 \]

### Step 9: Convert to angles
The angles corresponding to these sine values are:
- \( \sin \alpha = \frac{1}{2} \) corresponds to \( \alpha = \frac{\pi}{6} \)
- \( \sin \alpha = 1 \) corresponds to \( \alpha = \frac{\pi}{2} \)

### Conclusion
Thus, the interval for \( \alpha \) is:
\[ \left( \frac{\pi}{6}, \frac{\pi}{2} \right) \]