Factor: $\sin x (3-2\sin x)=0$
Now you have either $\sin x=0$ or $3-2\sin x=0$. The latter has no solutions over the reals. So the intersections occur at $x=0,\pi,2\pi $
Next plug in $\dfrac{\pi}{2} ,\dfrac{3\pi}{2} $ to figure out which function is bigger in each interval. This gives you
$\displaystyle A=\int_0^\pi(3\sin x-2\sin^2 x)dx+\int_\pi^{2\pi}(2\sin^2 x-3\sin x)dx $
As slip eternal said, the two curves, 3sin(x) and $2 sin^2(x)$ cross at x= 0, $x= \pi$, and $x= 2\pi$. He then said "Next, plug in $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ to figure out which function is bigger in each interval". When $x= \frac{\pi}{2}$, which is between 0 and $\pi$, $3sin(x)= 3sin(\pi/2)= 3$ while $2 sin^2(x)= 2 sin^2(\pi/2)= 2$. Since 3sin(x) is greater than $2sin^2(x)$ at $x= \frac{\pi}{2}$ (and both functions are continuous so the only place they can change order is where they are equal) $3sin(x)$ is larger than $2sin^2(x)$ for all x between 0 and $\pi$. So the area between them is $\int_0^{\pi} 3sin(x)- 2sin^2(x) dx$.
At $x= \frac{3\pi}{2}$, $3 sin(x)= 3sin(3\pi/2)= -3$ while $2sin^2(x)= 2 sin^2(3\pi/2)= 2$ so now $2 sin^2(x)$ is larger. The area between them is $\int_{\pi}^{2\pi} 2sin^2(x)- 3 sin(x) dx$. The total area is the sum of those two.