Trigonometric inequation $\sin x \ne \sin y$

How can I solve the following trigonometric inequation?

$$\sin\left(x\right)\ne \sin\left(y\right)\>,\>x,y\in \mathbb{R}$$

Why I'm asking this question... I was doing my calculus homework, trying to plot the domain of the function $f\left(x,y\right)=\frac{x-y}{sin\left(x\right)-sin\left(y\right)}$ and figured out I'd have to solve the inequation $\sin\left(x\right)\ne\sin\left(y\right)$... I was able to come to the answer $y\ne x +2\cdot k\cdot \pi \>,\>k \in \mathbb{N}$. However, the answer on the textbook also includes $y\ne -x +2\cdot k\cdot \pi + \pi \>,\>k \in \mathbb{N}$, so I thought that I was probably doing something wrong while solving that inequation.

• Have you tried solving first the equality? – An old man in the sea. Aug 14 '14 at 23:01
• $sin(x)$ is a $2pi$ periodic function so $sin(x) = sin(2pi+x)$. Try to use this fact in there. – Derek Orr Aug 14 '14 at 23:03
• @DerekOrr yes I think that's the main point. Just like $x^2 \neq y^2$ gives you $\pm x \neq \pm y$ not simply $x \neq y$. – TooTone Aug 14 '14 at 23:09
• Hi Frank, welcome to math SE. We like people who ask questions to provide some context as to why the question is difficult, what you have tried, etc. – TooTone Aug 14 '14 at 23:10
• @TooTone Ok, so let me explain why I'm asking this question... I was doing my calculus homework, trying to plot the domain of the function $f\left(x,y\right)=\frac{x-y}{sin\left(x\right)-sin\left(y\right)}$ and figured out I'd have to solve the inequation $\sin\left(x\right)\ne\sin\left(y\right)$... I was able to come to the answer $y\ne x +2\cdot k\cdot \pi \>,\>k \in \mathbb{N}$. However, the answer on the textbook also includes $y\ne -x +2\cdot k\cdot \pi + \pi \>,\>k \in \mathbb{N}$, so I thought that I was probably doing something wrong while solving that inequation. – Frank Aug 14 '14 at 23:27

The other part of it is: An angle and its supplement have the same sine, so you have to exclude angles that are supplementary to $x$ as well as those that are coterminal. $\pi - x$ is supplementary to $x$, and so $y = \pi - x + 2\pi k$ is a second set of solutions to $\sin y = \sin x$.
Let's ask the opposite question. For what values of $x,y$ is $f(x,y)$ not defined?
The only place where $f(x,y)$ is undefined is where $\sin x - \sin y = 0$.
Using the difference to product identities, we have $0 = \sin x - \sin y = 2\sin\dfrac{x-y}{2}\cos\dfrac{x+y}{2}$.
Thus, $f(x,y)$ is undefined iff either $\sin \dfrac{x-y}{2} = 0$ or $\cos\dfrac{x+y}{2} = 0$.
That will tell you where $f(x,y)$ isn't defined, and thus, tell you the domain of $f(x,y)$.