Solve sin(2x) dx + cos(3y) dy = 0, where y(pi/2) = pi/3

[5hassay]

Homework Statement

Solve the equation \sin(2x) dx + cos(3y) dy = 0, where y(\pi/2) = \pi/3

Homework Equations

N/A

The Attempt at a Solution

I understand the process that gets from the original equation to y = \frac{\arcsin(\frac{3}{2} (\cos(2x) + 1))}{3}

However, I don't understand why the answer provided is that but with a small change: y = \frac{\pi - \arcsin(\frac{3}{2} (\cos(2x) + 1))}{3}

I believe it has something to do with the properties of \sin, but I just can't see it.

Thank you

EDIT: Removed repeating problem, equations, attempt formats

 

[PhysicsGente]

When you integrate you obtain a constant term. For example,

\int{x}dx = \frac{x^2}{2} + C

Use the boundary condition to find it.

 

[5hassay]

When you integrate you obtain a constant term. For example,

\int{x}dx = \frac{x^2}{2} + C

Use the boundary condition to find it.

When you say "boundary condition," are you referring to how y(\pi/2) = \pi/3?

If so, I already found the constant C from the integration, namely C=\frac{1}{2}:

Integrating, I got \frac{1}{3} \sin(3y) = \frac{1}{2} \cos(2x) + C

Substituting, I find 0 = -\frac{1}{2} + C \Longrightarrow C = \frac{1}{2}

In solving, I got the equation in the OP

EDIT: Thanks for the reply