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?

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 $y=0$.
