Solutions of $\sin^2\theta = \frac{x^2+y^2}{2xy} $ If $x$ and $y$ are real, then the equation
$$\sin^2\theta = \frac{x^2+y^2}{2xy}$$
has a solution:


*

*for all $x$ and $y$  

*for no $x$ and $y$  

*only when $x \neq y \neq 0$  

*only when $x = y \neq 0$ 


First you need $x \ne 0$ and $y \ne 0$ in order for your fraction $\dfrac{x^2+y^2}{2xy}$ to make sense.
Then as $\sin \in [-1,1]$ you need $x^2+y^2 \leq 2xy$ which means:
$x^2 -2xy+y^2 \leq 0$ ie $(x-y)^2 \leq 0$
Therefore you have solutions on $\theta$ only if $x = y \ne 0$, and those solutions are $\theta = \dfrac{\pi}{2} \pmod \pi$
