solving an expression based on sin $\theta$ If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$, then $x$ must be equal to what?
What does the following solution mean?

$0 \le \sin^2 \le 1$
This implies $0 \le \frac{x^2 + y^2 + 1 }{ 2x } \le 1$
This implies $\frac{(x - 1)^2 + y^2 }{2x} \le 0 $
This implies $x = 1$.

Can you please explain me the solution?

On rearrangement we have $$x^2-2x\sin^2\theta+y^2+1=0\ \  \ \ (1)$$
which is clearly a Quadratic Equation in $x$
As $x$ is real, the discriminant must be $\ge0$
But actually  the discriminant, 
$\displaystyle (2\sin^2\theta)^2-4(y^2+1)=-4(y^2+1-\sin^4\theta)$
$\displaystyle=-4\{y^2+(1-\sin^2\theta)(1+\sin^2\theta)\}=-4\{y^2+\cos^2\theta(1+\sin^2\theta)\}$ which is $\not>0$ 
So for real $\displaystyle x, y$ and $\cos\theta$ must be individually equal to zero
$\displaystyle \implies \sin^2\theta=1-0^2=1$
So, $(1)$ reduces to  $\displaystyle x^2-2x+1=0$
