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.
Similar Questions
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Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
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- Let n be the number obtained on rolling a fair die. If the probability that the system
$x - ny + z = 6$
$x + (n - 2)y + (n + 1)z = 8$
$(n - 1)y + z = 1$
has a unique solution is $\frac{k}{6}$, then the sum of k and all possible values of n is : - If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
