Intersection of $y=a\sin2x$ and $y=\sin x$ Two graphs $y=a\sin2x$ and $y=\sin x$ are intersecting on $[0,\pi$].
Find a.
What I did:
$a\sin2x=\sin x$
$\sin x(2a\cos x-1)=0$
$\sin x=0$ (doesn't fit the criteria)
$2a\cos x-1=0$
$\cos x= \frac{1}{2a}$
In solutions says that a is $|a|> \frac{1}{2}$. How did they get that? Does anybody have any suggestions?

Note that $-1 \leq \cos x \leq 1$, so that we may write $-1 \leq \frac{1}{2a} \leq 1$.
Now we must be careful, since we're going to be multiplying these inequalities on both sides by $a$, but the direction of the inequality will flip if $a$ is negative. Let's consider two cases: first, if $a>0$, then from $\frac{1}{2a} \leq 1$ it is clear that $a \geq \frac{1}{2}$. On the other hand, if $a<0$, then $-1 \leq \frac{1}{2a}$ implies that $a \leq -\frac{1}{2}$.
Combining these two results, we have $|a| \geq \frac{1}{2}$.
