I tried to prove that $x(t)=A \sin(\omega t + \phi)$ knowing that $$\frac{\text{d}^2x}{\text{d}t^2}=-\omega^2 x$$ and $x=A$ when $v=0$, where $v = \frac{\text{d}x}{\text{d}t}$.
Since $\frac{\text{d}^2x}{\text{d}t^2} = v\frac{\text{d}v}{\text{d}x}$ I got the equation $$v \frac{\text{d}v}{\text{d}x}=-\omega^2 x \Rightarrow v\text{d} v = -\omega^2 x\text{d} x \Rightarrow v^2=\omega^2(A^2-x^2)$$
This is the point where I get confused. Taking square root gives
$$(1)\quad v=\pm \omega\sqrt{A^2-x^2} \Rightarrow \frac{\text{d}x}{\text{d}t} = \pm \omega \sqrt{A^2-x^2} $$
and the $\pm$ sign is annoying. Even though $\sqrt{A^2-x^2}$ is zero for some $t$s, I tried to write
$$(2) \quad \frac{dx}{\sqrt{A^2-x^2}}=\pm\omega dt \Rightarrow \arcsin\frac{x}{A}=\pm \omega t+ \phi \Rightarrow x(t) = A\sin(\pm \omega t +\phi)$$
Those are my two concerns: the $\pm$ sign in the final equation and the possible division by $0$ in $(2).$
How could I fix this?