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?

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
$$\sin^2\theta=\frac{x^2+1}{2x}=1\iff x=1$$
