The quadratic equation `(1 - sintheta) x^(2)+ 2(1 - sin theta) x - 3 sin theta = 0` has both roots complex for all `theta` lying in the interval

Options

A. `(-(pi)/2, (pi)/2)`
B. `(0,(3pi)/2)`
C. `((pi)/6, (7pi)/6)`
D. `((7pi)/6,(11pi)/6)`

Answer

To determine the interval of \( \theta \) for which the quadratic equation

\[ (1 - \sin \theta) x^2 + 2(1 - \sin \theta) x - 3 \sin \theta = 0 \]

has both roots complex, we need to analyze the discriminant of the equation. The roots of a quadratic equation are complex if the discriminant \( D \) is less than zero.

### Step 1: Identify the coefficients
The quadratic equation is in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 1 - \sin \theta \)
- \( b = 2(1 - \sin \theta) \)
- \( c = -3 \sin \theta \)

### Step 2: Calculate the discriminant
The discriminant \( D \) is given by:

\[ D = b^2 - 4ac \]

Substituting the values of \( a \), \( b \), and \( c \):

\[ D = [2(1 - \sin \theta)]^2 - 4(1 - \sin \theta)(-3 \sin \theta) \]

Calculating \( D \):

\[ D = 4(1 - \sin \theta)^2 + 12 \sin \theta (1 - \sin \theta) \]

### Step 3: Simplify the discriminant
Expanding the expression:

\[ D = 4(1 - 2\sin \theta + \sin^2 \theta) + 12 \sin \theta - 12 \sin^2 \theta \]

Combining like terms:

\[ D = 4 - 8\sin \theta + 4\sin^2 \theta + 12\sin \theta - 12\sin^2 \theta \]

\[ D = 4 + 4\sin \theta - 8\sin^2 \theta \]

### Step 4: Set the condition for complex roots
For the roots to be complex, we require:

\[ D < 0 \]

This gives us the inequality:

\[ 4 + 4\sin \theta - 8\sin^2 \theta < 0 \]

### Step 5: Rearranging the inequality
Rearranging the inequality:

\[ -8\sin^2 \theta + 4\sin \theta + 4 < 0 \]

Dividing through by -4 (and flipping the inequality):

\[ 2\sin^2 \theta - \sin \theta - 1 > 0 \]

### Step 6: Factor the quadratic
Factoring the quadratic:

\[ (2\sin \theta + 1)(\sin \theta - 1) > 0 \]

### Step 7: Determine the intervals
The critical points occur at:

\[ \sin \theta = -\frac{1}{2} \quad \text{and} \quad \sin \theta = 1 \]

### Step 8: Analyze the intervals
We need to find where the product is positive. The intervals to consider are:

1. \( \sin \theta < -\frac{1}{2} \)
2. \( -\frac{1}{2} < \sin \theta < 1 \)
3. \( \sin \theta > 1 \) (not possible)

The sine function is less than \(-\frac{1}{2}\) in the intervals:

\[ \frac{7\pi}{6} < \theta < \frac{11\pi}{6} \]

### Conclusion
Thus, the values of \( \theta \) for which the quadratic equation has both roots complex are:

\[ \theta \in \left( \frac{7\pi}{6}, \frac{11\pi}{6} \right) \]