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?

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.
