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