If `0 lt alpha lt beta lt gamma lt pi//2`, then the equation
`(x-sinbeta)(x-singamma)+(x-sinalpha)(x-singamma)+(x-sinalpha)(x-sinbeta)=0` has

Options

A. real and unequal roots
B. non-real roots
C. real and equal roots
D. real and unequal roots greater than `2`

Explanation

To solve the equation

\[ (x - \sin \beta)(x - \sin \gamma) + (x - \sin \alpha)(x - \sin \gamma) + (x - \sin \alpha)(x - \sin \beta) = 0, \]

we will first expand and simplify the left-hand side.

### Step 1: Expand Each Term
We will expand each of the three products in the equation.

1. **First term:**
\[ (x - \sin \beta)(x - \sin \gamma) = x^2 - x \sin \beta - x \sin \gamma + \sin \beta \sin \gamma \]

2. **Second term:**
\[ (x - \sin \alpha)(x - \sin \gamma) = x^2 - x \sin \alpha - x \sin \gamma + \sin \alpha \sin \gamma \]

3. **Third term:**
\[ (x - \sin \alpha)(x - \sin \beta) = x^2 - x \sin \alpha - x \sin \beta + \sin \alpha \sin \beta \]

### Step 2: Combine All Terms
Now, we will combine all the expanded terms:

\[ (x^2 - x \sin \beta - x \sin \gamma + \sin \beta \sin \gamma) + (x^2 - x \sin \alpha - x \sin \gamma + \sin \alpha \sin \gamma) + (x^2 - x \sin \alpha - x \sin \beta + \sin \alpha \sin \beta) = 0 \]

Combining like terms, we get:

- **Total \(x^2\) terms:** \(3x^2\)
- **Total \(x\) terms:** \(-x(\sin \alpha + \sin \beta + \sin \gamma) - x(\sin \alpha + \sin \beta + \sin \gamma) = -2x(\sin \alpha + \sin \beta + \sin \gamma)\)
- **Constant terms:** \(\sin \beta \sin \gamma + \sin \alpha \sin \gamma + \sin \alpha \sin \beta\)

Thus, the equation simplifies to:

\[ 3x^2 - 2x(\sin \alpha + \sin \beta + \sin \gamma) + (\sin \beta \sin \gamma + \sin \alpha \sin \gamma + \sin \alpha \sin \beta) = 0 \]

### Step 3: Identify Coefficients
This is a quadratic equation in the form \(ax^2 + bx + c = 0\), where:
- \(a = 3\)
- \(b = -2(\sin \alpha + \sin \beta + \sin \gamma)\)
- \(c = \sin \beta \sin \gamma + \sin \alpha \sin \gamma + \sin \alpha \sin \beta\)

### Step 4: Calculate the Discriminant
To determine the nature of the roots, we calculate the discriminant \(D\):

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

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

\[ D = [-2(\sin \alpha + \sin \beta + \sin \gamma)]^2 - 4 \cdot 3 \cdot (\sin \beta \sin \gamma + \sin \alpha \sin \gamma + \sin \alpha \sin \beta) \]

Calculating \(b^2\):

\[ D = 4(\sin \alpha + \sin \beta + \sin \gamma)^2 - 12(\sin \beta \sin \gamma + \sin \alpha \sin \gamma + \sin \alpha \sin \beta) \]

### Step 5: Analyze the Discriminant
Since \(\alpha, \beta, \gamma\) are all in the interval \((0, \frac{\pi}{2})\), the sine values are positive. Therefore, both \(b^2\) and \(c\) are positive.

### Conclusion
Since \(D > 0\), the roots of the quadratic equation are real and distinct.

### Final Answer
The equation has **real and distinct roots**.

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