Answer :
Certainly! Let's solve the equation [tex]\((x^2 + y^2) \sin \beta \cos \beta + xy = 0\)[/tex], and express [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex] and [tex]\(\beta\)[/tex].
### Step-by-Step Solution:
1. Original Equation:
[tex]\[ (x^2 + y^2) \sin \beta \cos \beta + xy = 0 \][/tex]
2. Rearrange to Isolate [tex]\(xy\)[/tex]:
[tex]\[ (x^2 + y^2) \sin \beta \cos \beta = -xy \][/tex]
3. Divide Both Sides by [tex]\(y\)[/tex]:
[tex]\[ \left(\frac{x^2}{y} + y\right) \sin \beta \cos \beta = -x \][/tex]
4. Substitute Example Values:
- Let [tex]\(y = 1\)[/tex] (an example value for [tex]\(y\)[/tex]).
- Let [tex]\(\beta = \frac{\pi}{4}\)[/tex] (an example value for [tex]\(\beta\)[/tex], which is 45 degrees).
Calculate [tex]\(\sin \beta\)[/tex] and [tex]\(\cos \beta\)[/tex]:
- [tex]\(\sin \beta = \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \approx 0.7071\)[/tex]
- [tex]\(\cos \beta = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \approx 0.7071\)[/tex]
5. Insert Calculated Values:
[tex]\[ \left(\frac{x^2}{1} + 1\right) \left(0.7071 \cdot 0.7071\right) = -x \][/tex]
Simplify:
[tex]\[ (x^2 + 1) \cdot 0.5 = -x \][/tex]
6. Rewrite in Quadratic Form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[ 0.5x^2 + x + 0.5 = 0 \][/tex]
Here:
- [tex]\(a = 0.5\)[/tex]
- [tex]\(b = 1\)[/tex]
- [tex]\(c = 0.5\)[/tex]
7. Find the Discriminant:
[tex]\[ \Delta = b^2 - 4ac = 1^2 - 4(0.5)(0.5) = 1 - 1 = 0 \][/tex]
8. Since the Discriminant is Zero:
The quadratic equation has exactly one real solution (a repeated root).
9. Use the Quadratic Formula [tex]\(x = \frac{-b \pm \sqrt{\Delta}}{2a}\)[/tex]:
[tex]\[ x = \frac{-1 \pm \sqrt{0}}{2 \cdot 0.5} = \frac{-1 \pm 0}{1} = -1 \][/tex]
### Summary:
- [tex]\(\sin \beta = 0.7071\)[/tex]
- [tex]\(\cos \beta = 0.7071\)[/tex]
- Coefficients: [tex]\(a = 0.5, b = 1, c = 0.5\)[/tex]
- Discriminant: [tex]\(\Delta = 0\)[/tex]
- Solutions for [tex]\(x\)[/tex]: [tex]\(x = -1.0\)[/tex] (repeated root)
Thus, the equation [tex]\((x^2 + y^2) \sin \beta \cos \beta + xy = 0\)[/tex] yields the solutions:
[tex]\[ x = -1.0 \][/tex]
This detailed solution provides the derived values step-by-step.
### Step-by-Step Solution:
1. Original Equation:
[tex]\[ (x^2 + y^2) \sin \beta \cos \beta + xy = 0 \][/tex]
2. Rearrange to Isolate [tex]\(xy\)[/tex]:
[tex]\[ (x^2 + y^2) \sin \beta \cos \beta = -xy \][/tex]
3. Divide Both Sides by [tex]\(y\)[/tex]:
[tex]\[ \left(\frac{x^2}{y} + y\right) \sin \beta \cos \beta = -x \][/tex]
4. Substitute Example Values:
- Let [tex]\(y = 1\)[/tex] (an example value for [tex]\(y\)[/tex]).
- Let [tex]\(\beta = \frac{\pi}{4}\)[/tex] (an example value for [tex]\(\beta\)[/tex], which is 45 degrees).
Calculate [tex]\(\sin \beta\)[/tex] and [tex]\(\cos \beta\)[/tex]:
- [tex]\(\sin \beta = \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \approx 0.7071\)[/tex]
- [tex]\(\cos \beta = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \approx 0.7071\)[/tex]
5. Insert Calculated Values:
[tex]\[ \left(\frac{x^2}{1} + 1\right) \left(0.7071 \cdot 0.7071\right) = -x \][/tex]
Simplify:
[tex]\[ (x^2 + 1) \cdot 0.5 = -x \][/tex]
6. Rewrite in Quadratic Form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[ 0.5x^2 + x + 0.5 = 0 \][/tex]
Here:
- [tex]\(a = 0.5\)[/tex]
- [tex]\(b = 1\)[/tex]
- [tex]\(c = 0.5\)[/tex]
7. Find the Discriminant:
[tex]\[ \Delta = b^2 - 4ac = 1^2 - 4(0.5)(0.5) = 1 - 1 = 0 \][/tex]
8. Since the Discriminant is Zero:
The quadratic equation has exactly one real solution (a repeated root).
9. Use the Quadratic Formula [tex]\(x = \frac{-b \pm \sqrt{\Delta}}{2a}\)[/tex]:
[tex]\[ x = \frac{-1 \pm \sqrt{0}}{2 \cdot 0.5} = \frac{-1 \pm 0}{1} = -1 \][/tex]
### Summary:
- [tex]\(\sin \beta = 0.7071\)[/tex]
- [tex]\(\cos \beta = 0.7071\)[/tex]
- Coefficients: [tex]\(a = 0.5, b = 1, c = 0.5\)[/tex]
- Discriminant: [tex]\(\Delta = 0\)[/tex]
- Solutions for [tex]\(x\)[/tex]: [tex]\(x = -1.0\)[/tex] (repeated root)
Thus, the equation [tex]\((x^2 + y^2) \sin \beta \cos \beta + xy = 0\)[/tex] yields the solutions:
[tex]\[ x = -1.0 \][/tex]
This detailed solution provides the derived values step-by-step.