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?

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.
