Intersection of $y=a\sin2x$ and $y=\sin x$

Question

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?

Answer 1

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 $|a| \geq \frac{1}{2}$.

Answer 2

It's simple if you remember that $|\cos x|\le 1$, so you must have $$ \frac{1}{2|a|}\le1 $$ which is equivalent to $$ |a|\ge\frac{1}{2}. $$ You can multiply both sides by $|a|$ because it's a positive number (it can't obviously be $0$).