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?
 A: Certainly, for any value of $\theta$, we have $-1 \leq \sin \theta \leq 1$, and therefore, $0 \leq \sin^2 \theta \leq 1$, which bounds the left-hand side of your equation
$$
\sin^2 \theta = \frac{x^2+y^2+1}{2x}.
$$
Since the left-hand side is bounded, so must be the right-hand side, since they are equal. Thus,
$$
0 \leq \frac{x^2+y^2+1}{2x} \leq 1.
$$
Assume $x>0$ and multiply both sides by $2x$, you get
$$
0 \leq x^2+y^2+1\leq 2x
$$
or if you consider the rght-hand side only:
$$
0 = 2x - 2x \geq x^2+y^2+1-2x = \left(x^2-2x+1\right)+y^2 = (x-1)^2+y^2.
$$
Thus, $(x-1)^2+y^2 \leq 0$. But how can a sum of two squares be zero -- both squares are non-negative?!! This only happens if both are zero by themselves. So $y^2 = 0$ (which implies $y=0$) and $(x-1)^2 = 0$ (which implies $x=1$).
Notice $x\neq 0$ otherwise you cannot divide by $2x$ and the problem is invalid. If $x < 0$ on the other hand, the inequality flips:
$$
0 \geq x^2 + y^2 + 1 \geq 2x
$$
but the left-hand side makes no sense, since for any real $x,y$, we must have $x^2+y^2+1 \geq 1 > 0$. So $x$ cannot be negative.
Thus the only acceptable solution is $x=1$ and $y=0$.
A: We have $$(x-1)^2=x^2+1-2x\ge0\iff x^2+1\ge 2x$$
and since $\sin^2\theta\ge0$ hence we have $x>0$ and then
$$1\ge\sin^2\theta =\frac{x^2+y^2+1}{2x}\ge1\quad (>1\;\text{if}\; y\ne 0\;\text{which gives a contradiction})$$
hence $y=0$ and
$$\sin^2\theta=\frac{x^2+1}{2x}=1\iff x=1$$
A: Well...
The range of $\sin^2$ is $[0,1]$. That is the first step.
Now, $\frac{x^2+y^2+1}{2x} \leq 1$ implies $\frac{x^2+y^2+1-2x}{2x} \leq 0$. Now, rearranging, $\frac{x^2-2x+1+y^2}{2x} = \frac{(x-1)^2+y^2}{2x} \leq 0$, which is true if and only if the numerator is zero or $x < 0$. Now $x>0$, since otherwise $\sin^2(x) < 0$ and this is outside the acceptable range. So, $(x-1)^2+y^2 = 0 \iff x=1,y=0$. Hence, $x=1$ as desired.
A: 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  $\displaystyle x^2-2x+1=0$
