Sure, let's solve the equation [tex]\(\left(1+\sin^2 \alpha\right) x^2-2 x y+y^2 \cos^2 \alpha=0\)[/tex] step-by-step for [tex]\(x\)[/tex].
1. Identify the equation:
[tex]\[
(1 + \sin^2(\alpha)) x^2 - 2xy + y^2 \cos^2(\alpha) = 0
\][/tex]
2. Attempt to solve for [tex]\(x\)[/tex]:
This is a quadratic equation in [tex]\(x\)[/tex] of the form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[
a = 1 + \sin^2(\alpha), \quad b = -2y, \quad c = y^2 \cos^2(\alpha)
\][/tex]
3. Quadratic formula:
The general solution for a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
4. Plug in the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
x = \frac{-(-2y) \pm \sqrt{(-2y)^2 - 4(1 + \sin^2(\alpha))(y^2 \cos^2(\alpha))}}{2(1 + \sin^2(\alpha))}
\][/tex]
5. Simplify the terms under the square root:
[tex]\[
x = \frac{2y \pm \sqrt{4y^2 - 4(1 + \sin^2(\alpha))(y^2 \cos^2(\alpha))}}{2(1 + \sin^2(\alpha))}
\][/tex]
[tex]\[
x = \frac{2y \pm \sqrt{4y^2 - 4y^2(1 + \sin^2(\alpha))\cos^2(\alpha)}}{2(1 + \sin^2(\alpha))}
\][/tex]
[tex]\[
x = \frac{2y \pm \sqrt{4y^2 (1 - (1 + \sin^2(\alpha))\cos^2(\alpha))}}{2(1 + \sin^2(\alpha))}
\][/tex]
6. Factor out the common term under the square root:
[tex]\[
x = \frac{2y \pm \sqrt{4y^2 (1 - \cos^2(\alpha) - \cos^2(\alpha)\sin^2(\alpha))}}{2(1 + \sin^2(\alpha))}
\][/tex]
Recall the Pythagorean identity: [tex]\(1 - \cos^2(\alpha) = \sin^2(\alpha)\)[/tex], so:
[tex]\[
x = \frac{2y \pm \sqrt{4y^2 (\sin^2(\alpha) - \cos^2(\alpha)\sin^2(\alpha))}}{2(1 + \sin^2(\alpha))}
\][/tex]
[tex]\[
x = \frac{2y \pm \sqrt{4y^2 \sin^2(\alpha)(1 - \cos^2(\alpha))}}{2(1 + \sin^2(\alpha))}
\][/tex]
Again using the identity [tex]\(1 - \cos^2(\alpha) = \sin^2(\alpha)\)[/tex], we have:
[tex]\[
x = \frac{2y \pm \sqrt{4y^2 \sin^2(\alpha) \cdot \sin^2(\alpha)}}{2(1 + \sin^2(\alpha))}
\][/tex]
[tex]\[
x = \frac{2y \pm \sqrt{4y^2 \sin^4(\alpha)}}{2(1 + \sin^2(\alpha))}
\][/tex]
[tex]\[
x = \frac{2y \pm 2y\sqrt{\sin^4(\alpha)}}{2(1 + \sin^2(\alpha))}
\][/tex]
[tex]\[
x = \frac{2y (1 \pm \sqrt{\sin^4(\alpha)})}{2(1 + \sin^2(\alpha))}
\][/tex]
[tex]\[
x = \frac{y (1 \pm \sqrt{\sin^4(\alpha)})}{1 + \sin^2(\alpha)}
\][/tex]
7. Combine the terms:
Therefore, the solutions are:
[tex]\[
x = \frac{y (1 - \sqrt{\sin^4(\alpha)})}{1 + \sin^2(\alpha)}
\][/tex]
and
[tex]\[
x = \frac{y (1 + \sqrt{\sin^4(\alpha)})}{1 + \sin^2(\alpha)}
\][/tex]
In summary, the equation [tex]\((1+\sin ^2 \alpha) x^2-2 x y+y^2 \cos ^2 \alpha=0\)[/tex] has the solutions:
[tex]\[
x = \frac{y (1 - \sqrt{\sin^4(\alpha)})}{1 + \sin^2(\alpha)}
\][/tex]
and
[tex]\[
x = \frac{y (1 + \sqrt{\sin^4(\alpha)})}{1 + \sin^2(\alpha)}
\][/tex]
