Easiest way to solve $\frac{2}{9}x=\sin(\frac{\pi}{9}x)$ We wanted to find the points of intersection between 2 curves, so we have to solve this equation $\frac{2}{9} x= \sin(\frac{\pi}{9}x)$.
I couldn't figure out an easy way to find the points of intersection, they are $-\frac{9}{2}$, $0$, and $\frac{9}{2}$.
Any ideas please on how to solve such equations?
 A: What tools are you allowed to use to solve this problem? There are multiple ways to solve this based on your background. If it's just basic algebra and trig identities, list all the values of $x$ for which $\sin(\frac{\pi}{9}x) = c$ where $c$ are the known sin values for special angles. Then it's just a matter of plugging in those values for the LHS and see if it's equal. Now you know that sin only gives a value between -1 and 1, so your LHS will also be restricted so x values can be restricted very quickly.
So $|\frac{2}{9}x| \leq 1$, then $|x| \leq 9/2$. So you'll only have to test angles between $0$ and $\pm \frac{\pi}{2}$ since $\frac{\pi}{9}(\frac{9}{2})=\frac{\pi}{2}$
A: uhm the following way isn't that easy... you have to consider several properties of functions...
first off: $\sin(\dfrac{\pi x}{9}) = \dfrac{2x}{9} \iff \sin^{-1}(\dfrac{2x}{9}) = \dfrac{\pi x}{9}$
Let: $f(x) =\sin^{-1}(\dfrac{2x}{9}) - \dfrac{\pi x}{9} $
Our goal is to find where $f(x) = 0$
Notice $f(x) = -f(-x) \implies f(x)$ is an odd function. If an odd function is defined at zero, then its graph must pass through the origin. Proof. This means $x=0$ is a solution.
Now lets examine the domain of $f(x)$. Note $\sin^{-1}(x)$ is defined only when $-1\leq x\leq1$
This means domain of $f(x) =[-\dfrac{9}{2},\dfrac{9}{2}]$
Now lets examine the derivative of f(x):
$f'(x) = \dfrac{2}{9\sqrt{1-\dfrac{4x^2}{81}}} - \dfrac{\pi}{9}$
Now $f'(x) =0 \implies x = \pm \dfrac{9\sqrt{1-\dfrac{4}{\pi^2}}}{2} = \pm a, a>0$.
After drawing a sign diagram one can see:
f(x) is increasing $-\dfrac{9}{2}\leq x\leq -a$
f(x) is decreasing $-a\leq x\leq a$
f(x) is increasing $a\leq x\leq \dfrac{9}{2}$
Since f(x) is strictly decreasing in the interval $[-a,a]$ where $a>0$ and $f(-a) >0$ and $f(a)<0$, then there must be only one root from $[-a,a]$ and we already saw that $x = 0$ is the solution and now we can confirm that this is the only root in $[-a,a]$ Also notice the boundary points are $\pm \dfrac{9}{2}$ where the function is zero. With the same increasing/decreasing function logic we can confirm these two are also the only roots in their respective interval.
