To solve the equation [tex]\( x^2 \sin^2 \lambda - 2xy + y^2 \left( 1 + \cos^2 \lambda \right) = 0 \)[/tex], let's take a step-by-step approach:
1. Understand the Equation: The equation we are given is:
[tex]\[
x^2 \sin^2 \lambda - 2xy + y^2 \left( 1 + \cos^2 \lambda \right) = 0
\][/tex]
This is a quadratic equation in terms of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] with trigonometric functions involving [tex]\( \lambda \)[/tex].
2. Examine the Terms Individually: The terms in the equation include:
- [tex]\( x^2 \sin^2 \lambda \)[/tex]
- [tex]\( -2xy \)[/tex]
- [tex]\( y^2 \left( 1 + \cos^2 \lambda \right) \)[/tex]
3. Combine Like Terms:
Combining these terms in the equation gives a quadratic polynomial form:
[tex]\[
x^2 \sin^2 \lambda - 2xy + y^2 + y^2 \cos^2 \lambda = 0
\][/tex]
4. Simplifying the Equation:
Notice that [tex]\( y^2 \left( 1 + \cos^2 \lambda \right) \)[/tex] can be rewritten as:
[tex]\[
y^2 + y^2 \cos^2 \lambda
\][/tex]
5. Rewriting the Equation:
Putting it all together, we rewrite the equation as:
[tex]\[
x^2 \sin^2 \lambda - 2xy + y^2 + y^2 \cos^2 \lambda = 0
\][/tex]
Simplifying further, it remains as:
[tex]\[
x^2 \sin^2 \lambda - 2xy + y^2 (1 + \cos^2 \lambda) = 0
\][/tex]
6. Confirmation:
This matches our given initial equation:
[tex]\[
x^2 \sin^2 \lambda - 2xy + y^2 (1 + \cos^2 \lambda) = 0
\][/tex]
So, the detailed, step-by-step solution results in confirming:
[tex]\[
\boxed{ x^2 \sin^2 \lambda - 2xy + y^2 \left( 1 + \cos^2 \lambda \right) = 0 }
\][/tex]
This verifies that the equation is consistent and correct.
