If equation `2x^(3)-6x+2sina+3=0,a in (0,pi)` has only one real root, then the largest interval in which a lies is

Options

A. `(0,(pi)/(6))`
B. `((pi)/(6),(pi)/(3))`
C. `((pi)/(6),(5pi)/(6))`
D. `((5pi)/(6),pi)`

Explanation

To solve the problem, we need to analyze the function given by the equation \( f(x) = 2x^3 - 6x + 2\sin a + 3 \) and determine the conditions under which it has only one real root.

### Step 1: Define the function
We define the function as:
\[ f(x) = 2x^3 - 6x + 2\sin a + 3 \]

### Step 2: Find the derivative
Next, we find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3 - 6x + 2\sin a + 3) = 6x^2 - 6 \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ 6x^2 - 6 = 0 \]
\[ x^2 - 1 = 0 \]
\[ x = \pm 1 \]

### Step 4: Analyze the critical points
We have critical points at \( x = -1 \) and \( x = 1 \). To determine the nature of these points, we can evaluate the function at these points.

### Step 5: Evaluate \( f(-1) \) and \( f(1) \)
Now we evaluate the function at these critical points:
\[ f(-1) = 2(-1)^3 - 6(-1) + 2\sin a + 3 = -2 + 6 + 2\sin a + 3 = 7 + 2\sin a \]
\[ f(1) = 2(1)^3 - 6(1) + 2\sin a + 3 = 2 - 6 + 2\sin a + 3 = -1 + 2\sin a \]

### Step 6: Determine conditions for one real root
For the function \( f(x) \) to have only one real root, it must either touch the x-axis at one of the critical points or be tangent to it. This means that one of the values must be zero while the other must be positive.

1. **Condition 1**: \( f(-1) = 0 \)
\[ 7 + 2\sin a = 0 \implies \sin a = -\frac{7}{2} \quad \text{(not possible since } \sin a \text{ must be between -1 and 1)} \]

2. **Condition 2**: \( f(1) = 0 \)
\[ -1 + 2\sin a = 0 \implies \sin a = \frac{1}{2} \]
This gives us \( a = \frac{\pi}{6} \) or \( a = \frac{5\pi}{6} \).

### Step 7: Determine the intervals
Now we need to check the sign of \( f(-1) \) and \( f(1) \):
- For \( f(-1) > 0 \):
\[ 7 + 2\sin a > 0 \implies \sin a > -\frac{7}{2} \quad \text{(always true)} \]

- For \( f(1) > 0 \):
\[ -1 + 2\sin a > 0 \implies \sin a > \frac{1}{2} \]
This implies:
\[ a \in \left(\frac{\pi}{6}, \frac{5\pi}{6}\right) \]

### Conclusion
Thus, the largest interval in which \( a \) lies such that the equation has only one real root is:
\[ \boxed{\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)} \]