First order differential equation, separable 
$g(x)$ is continuous on $[1,2]$ such that $g(1)=0$ and $$g'(x)x^2=\sqrt{1-g^2(x)}$$
  Find $g(2)$.

I found that $g(x)=\sin(c-\frac{1}{x})$ and since $g(1) = 0$ shouldn't my $c$ be equal to 1, so that $\sin(1-1)=0$.
But when I try for $g(2)$ with $c=1$, I am getting a different answer from the textbook.

Your solution, I mean one parameter family of solutions, are correct. Indeed $$g(y)=\sin\left(\frac{-1}x+c\right)$$ Now if we are given $g(1)=0$ so $$0=g(1)=\sin\left(\frac{-1}1+c\right)=\sin(c-1)$$ and then $c=1+k\pi,~~~k\in\mathbb Z$. Hence $$g(2)=\sin(-1/2+c)=\sin(1/2+k\pi),~~~k\in\mathbb Z$$
