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 $x = y \neq 0$ 


HINT:
$$(x/y)^2-2(x/y)\sin^2\theta+1=0$$
As $x/y$ is real,the discriminant is $$(2\sin^2\theta)^2-4\ge0\iff\sin^4\theta\ge1\iff\sin^2\theta\ge1\iff\cos^2\theta\le0$$
this is only possible if $\cos^2\theta=0\iff\sin^2\theta=?$
Can you take it from here?
