Step 1: Understanding the Concept:
This problem can be solved using the properties of rational functions and the Intermediate Value Theorem (IVT) or by considering the derivative of a cubic polynomial. The equation given is of the form \( \sum \frac{1}{x-a_i} = 0 \).
Step 2: Key Formula or Approach:
Let \( f(x) = (x-\sin \alpha)(x-\sin \beta) + (x-\sin \beta)(x-\sin \gamma) + (x-\sin \gamma)(x-\sin \alpha) \).
This is a quadratic equation obtained by clearing the denominators of the original equation. We will evaluate this expression at specific points to check for sign changes.
Step 3: Detailed Explanation:
Let \( a = \sin \alpha \), \( b = \sin \beta \), and \( c = \sin \gamma \). Since \( 0<\alpha<\beta<\gamma<\pi/2 \), we know that \( 0<a<b<c<1 \).
The equation is:
\[ \frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = 0 \]
This is equivalent to:
\[ P(x) = (x-b)(x-c) + (x-a)(x-c) + (x-a)(b-x) = 0 \]
(Wait, the signs: \( P(x) = (x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) \))
Now, let's evaluate \( P(x) \) at the points \( a, b, \) and \( c \):
- \( P(a) = (a-b)(a-c) + 0 + 0 \). Since \( a<b \) and \( a<c \), both factors are negative. Their product is positive: \( P(a)>0 \).
- \( P(b) = 0 + (b-a)(b-c) + 0 \). Since \( b>a \) (positive) and \( b<c \) (negative), their product is negative: \( P(b)<0 \).
- \( P(c) = 0 + 0 + (c-a)(c-b) \). Since \( c>a \) and \( c>b \), both factors are positive. Their product is positive: \( P(c)>0 \).
By the Intermediate Value Theorem:
1. Since \( P(a)>0 \) and \( P(b)<0 \), there must be at least one real root in the interval \( (a, b) \).
2. Since \( P(b)<0 \) and \( P(c)>0 \), there must be at least one real root in the interval \( (b, c) \).
As \( P(x) \) is a quadratic equation (degree 2), it can have at most 2 roots. We have found two distinct real roots in two non-overlapping intervals.
Step 4: Final Answer:
The roots are real and lie in the intervals \( (\sin \alpha, \sin \beta) \) and \( (\sin \beta, \sin \gamma) \). Thus, they are real and unequal.