Find the particular solution of $\sin(2x)dx + \cos(3y)dy = 0$ for $y(\pi/2) = \pi/3$ Find the particular solution of $\sin(2x)dx + \cos(3y)dy = 0$ for $y(\pi/2) = \pi/3$, plot the graph of the solution, and determine the interval in which the solution is defined.
Because the terms are separable, we integrate the expression $\sin(2x)dx = -\cos(3y)dy$ to yield $$-\frac{1}{2}\cos(2x) = -\frac{1}{3}\sin(3y) + C$$ We plug in our initial conditions and simplify as follows $$-\frac{1}{2}(-1) = -\frac{1}{3}(0) + C \rightarrow C = \frac{1}{2}$$ Thus, our implicit particular solution is of the form $-\cos(2x)/2 = -\sin(3y)/3 + 1/2$. Explicitly, this is $$y =\frac{\arcsin(\frac{3\cos(2x)+3}{2})}{3}$$ Utilizing the trigonometric identity $\cos^2(x) = 1/2 + \cos(2x)/2$, we simplify the above expression to $$y = \frac{\arcsin(3\cos^2(x))}{3}$$ However, this is not the correct answer. When we plug in our initial condition $x = \pi/2$, we get $0$ and not $\pi/3$. To accommodate for this shortcoming, we must look at the behavior of the function above. We note that $3\cos^2(2x) \leq 1$, because the domain of $\arcsin x$ is $0 \leq x \leq 1$.
That's all I got. This guy (http://www.math.drexel.edu/~tolya/example1.pdf) solves the problem, but beyond what I wrote above, I don't know where to go with it. Plotting the graph would be easy once solved, but I'm hooked up at the interval of definition as well.
Any help would be much appreciated.

Your calculus is correct. 
$$-\frac{1}{2}\cos(2x) = -\frac{1}{3}\sin(3y) + C$$
The condition $\quad y(\pi/2) = \pi/3\quad $ leads to $\quad C=\frac{1}{2}$
Note that, if the condition was $\quad y(\pi/2) = k\pi/3\quad $ the constant would be the same $\quad C=\frac{1}{2}\quad$ because $\quad \sin(3k\pi/3)=\sin(3\pi/3)=0$.
Thus, for a particular value of the constant $\quad C=\frac{1}{2}\quad$ the solution of the ODE is not only one particular function :
$$y =\frac{\arcsin(\frac{3\cos(2x)+3}{2})}{3} = \frac{\arcsin(3\cos^2(x))}{3}$$
but is a family of functions :
$$y =\frac{\arcsin(\frac{3\cos(2x)+3}{2})}{3} +\frac{k\pi}{3} = \frac{\arcsin(3\cos^2(x))}{3}+\frac{k\pi}{3}$$
where $k$ is any integer.
The particular solution according to the condition $\quad y(\pi/2) = \pi/3\quad $ has to be defined among this family of solutions, with a determined value of $k$, in the present case $k=1$.
$$y(x) =\frac{\arcsin(\frac{3\cos(2x)+3}{2})+\pi}{3}=\frac{\arcsin(3\cos^2(x))+\pi}{3}$$
