Trigonometric Equation Root Analysis

We are given the equation: $ \frac{1}{x - \sin \alpha} + \frac{1}{x - \sin \beta} + \frac{1}{x - \sin \gamma} = 0 $ with the condition $0 < \alpha < \beta < \gamma < \frac{\pi}{2}$.

Equation Simplification Steps

Let $a = \sin \alpha$, $b = \sin \beta$, and $c = \sin \gamma$. The condition $0 < \alpha < \beta < \gamma < \frac{\pi}{2}$ implies that $a, b, c$ are distinct real numbers satisfying $0 < a < b < c < 1$. The equation transforms to: $ \frac{1}{x - a} + \frac{1}{x - b} + \frac{1}{x - c} = 0 $ To simplify, combine the fractions by multiplying with the common denominator $(x - a)(x - b)(x - c)$, assuming $x \neq a, x \neq b, x \neq c$:

$ (x - b)(x - c) + (x - a)(x - c) + (x - a)(x - b) = 0 $

Expand the products:

$ (x^2 - (b+c)x + bc) + (x^2 - (a+c)x + ac) + (x^2 - (a+b)x + ab) = 0 $

Group terms to form a quadratic equation:

$ 3x^2 - 2(a+b+c)x + (ab+ac+bc) = 0 $

Discriminant Calculation and Analysis

This is a quadratic equation $Ax^2 + Bx + C = 0$, where $A=3$, $B=-2(a+b+c)$, and $C=ab+ac+bc$. We examine the discriminant, $\Delta = B^2 - 4AC$, to determine the nature of the roots.

$ \Delta = (-2(a+b+c))^2 - 4(3)(ab+ac+bc) $ $ \Delta = 4(a+b+c)^2 - 12(ab+ac+bc) $ $ \Delta = 4(a^2+b^2+c^2 + 2ab+2ac+2bc) - 12(ab+ac+bc) $ $ \Delta = 4(a^2+b^2+c^2 - ab-ac-bc) $

This expression for $\Delta$ can be rewritten as:

$ \Delta = 2 [ (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ac + a^2) ] $ $ \Delta = 2 [ (a-b)^2 + (b-c)^2 + (c-a)^2 ] $

Root Nature Based on Sine Conditions

Given $0 < a < b < c < 1$, the values $a, b, c$ are distinct.

• Since $a, b, c$ are distinct, $(a-b)^2$, $(b-c)^2$, and $(c-a)^2$ are all positive.
• Consequently, the discriminant $\Delta = 2 [ (a-b)^2 + (b-c)^2 + (c-a)^2 ]$ is strictly positive ($\Delta > 0$).
• A positive discriminant signifies that the quadratic equation has two distinct real roots.

These roots are guaranteed not to be equal to $a, b,$ or $c$. If $x=a$ were a root, substituting into the quadratic equation would yield $(a-b)(a-c)=0$, which is impossible since $a, b, c$ are distinct. The same logic applies to $x=b$ and $x=c$. Thus, the two distinct real roots found are the valid solutions to the original equation.

Conclusion on Root Type

The equation has real and unequal roots.