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 $x = y \neq 0$ 

 A: 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 $\theta = \dfrac{\pi}{2} \pmod \pi$
A: 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?
A: Suppose that $x$ and $y$ are not $0$.
We know that $$x^2 + y^2 \geq 2xy. \quad (1)$$
Two cases:


*

*$x>0$ and $y<0$ (or $x<0$ and $y>0$), then $\frac{x^2+y^2}{2xy} < 0$, and you cannot have $\sin^2(\theta) < 0$

*$x>0$ and $y>0$ (or $x<0$ and $y<0$). Then $(1)$ becomes 
$$\frac{x^2+y^2}{2xy} \geq 1.$$
The only solution is to have $\frac{x^2+y^2}{2xy} = 1$.
A: $$\begin{align}
  & {{\sin }^{2}}\theta =\frac{1}{2}\left( \frac{x}{y}+\frac{y}{x} \right)\ge 1\,\,\,\,\quad,\,\,\,\,\,\,\,\,\,\frac{x}{y}>0\,\,\,\,\,\,\Rightarrow \,\,\,\,\theta =k\pi +\frac{\pi }{2}\, \\ 
 & {{\sin }^{2}}\theta =\frac{1}{2}\left( \frac{x}{y}+\frac{y}{x} \right)\le -1\,\,\,\,\,,\,\,\,\frac{x}{y}<0\,\,\,\,\,\, \\ 
\end{align}$$ 
