If roots of the equation `2x^2-4x+2sintheta-1=0` are of opposite sign, then `theta` belongs

Options

A. `(pi/6, (5pi)/6)`
B. `(0, pi/6) uu ((5pi)/6, 2pi)`
C. `((13 pi)/6, (17 pi)/6)`
D. `(0, pi)`

Explanation

To solve the problem, we need to determine the range of \(\theta\) for which the roots of the quadratic equation \(2x^2 - 4x + 2\sin\theta - 1 = 0\) are of opposite signs.

### Step-by-Step Solution:

1. **Identify the coefficients of the quadratic equation**:
The given equation is \(2x^2 - 4x + (2\sin\theta - 1) = 0\). Here, the coefficients are:
- \(a = 2\)
- \(b = -4\)
- \(c = 2\sin\theta - 1\)

2. **Condition for roots to be of opposite signs**:
For the roots of a quadratic equation \(ax^2 + bx + c = 0\) to be of opposite signs, the product of the roots must be negative. The product of the roots is given by \(\frac{c}{a}\). Therefore, we need:
\[ \frac{c}{a} < 0 \implies \frac{2\sin\theta - 1}{2} < 0 \]

3. **Simplify the inequality**:
This simplifies to:
\[ 2\sin\theta - 1 < 0 \implies 2\sin\theta < 1 \implies \sin\theta < \frac{1}{2} \]

4. **Determine the range of \(\theta\)**:
The sine function is less than \(\frac{1}{2}\) in the intervals:
- From \(0\) to \(\frac{\pi}{6}\)
- From \(\frac{5\pi}{6}\) to \(\pi\)

Therefore, the solution for \(\theta\) is:
\[ \theta \in \left(0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, \pi\right) \]

### Final Answer:
\[ \theta \text{ belongs to } \left(0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, \pi\right) \]