How do I find the $y$ in $y^{\sin x} = x^{\sin y}$? I know how to derivate and I've found the implicit differentiation of $y^{\sin x} = x^{\sin y}$ which is $y' = \frac{\frac{\sin x}{y} - \cos x \ln y}{\frac{\sin y}{x} - \cos y \ln x}$, but how do I obtain $y$ alone, is there a way?
PD: Thanks you all for helping me understand this problem :)
 A: Apparently $\sin x=0$ or $\sin y=0$ doesn't satisfy the equation.
And $x=y>0$ are solutions to the equation. We now only consider solutions in which $x\ne y$.
Assuming $x,y>0$, take logarithm on both side, we have
$\sin x \ln y = \sin y \ln x$
or 
$\cfrac {\ln x}{\sin x}=\cfrac{\ln y}{\sin y}$
Define $f(x)=\cfrac {\ln x}{\sin x}$ 
Plot the graph of f(x), test it with $y=c$ ($c \in \mathbb{R}$), provided the graph intersect $y=c$ for more than once, the corresponding $x$ values of the intersections form a set $\{x_i\}, i\ge 2$, $x=x_i , y=x_j, i\ne j$ is a solution to the original equation with $x\ne y$. 

PS: for example, the line $y=10$ will intersect the graph of our $f(x)$ over many dozens of times, let's say it's $k$ times; name the x-coordinates of these intersection $x_1, x_2, \cdots, x_k$. Apparently $f(x_1)=f(x_2)=\cdots=f(x_k)$, which implies $\cfrac{\ln x_1}{\sin x_1}=\cfrac{\ln x_2}{\sin x_2}=\cdots=\cfrac{\ln x_k}{\sin x_k}$ which in turn means $x=x_1, y=x_2$, etc (Pick $2$ out of $k$ and assign them to $x,y$) are all solutions to the original equation. You will have to find the exact values numerically though.
A: If you plot the $(x,y)$ pairs for which this holds, you'll see that this won't only be problematic for $x=0$
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A: By symmetry, the equation $y^{\sin x} = x^{\sin y}$ is satisfied if 
$$y=x.$$ 
EDIT:
There might be more solutions, for instance:
If $x=0$, the equation for $y$ is $y^0=1=0^{\sin y}$ which holds for $\sin y = 0$ and thus $y = \pi n$ with $n\in\mathbb{Z}$.
A: It is the same as 
$$x^{1/\sin(x)}=y^{1/\sin(y)}$$
You can plot $x^{1/\sin x}$ on Wolfram Alpha.
That function seems to be increasing between $0$ and $\pi$, so there are no solutions below $\pi$, but there seem to be solutions between $0$ and $2\pi$.
A: We'll only consider the equation when $x$ and $y$ are both positive. (The expression $x^{\sin y}$ is, in general, not well defined when $x$ is negative, and likewise for $y^{\sin x}$.)
For $x,y\gt0$, the equation is clearly satisfied if $\sin x=\sin y=0$, since both sides become equal to $1$.  Thus the solution set includes the set $\{(m\pi,n\pi)\mid m,n\in\mathbb{N}\}$.
If $\sin x\sin y\not=0$, then $x^{\sin y}=y^{\sin x}$ is equivalent to $x^{1/\sin x}=y^{1/\sin y}$.  Setting aside the interval $(0,\pi)$, it's easy to see that, on any interval of the form $(2m\pi,(2m+1)\pi)$, the function $f(x)=x^{1/\sin x}$ varies continuously from $\infty$ down to $(2m+1/2)\pi$ and then back up to $\infty$, while on intervals of the form $((2m-1)\pi,2m\pi)$, it varies continuously from $0$ up to $((2m-1/2)\pi)^{-1}$ and then back down to $0$. (Note, the function doesn't actually take the values $\infty$ or $0$; those are just the appropriate limits at the endpoints.) Consequently, almost every value in the range of $f$ is achieved multiple times (and the closer the value is to $\infty$ or $0$, the more often it is achieved). This implies a continuum of solutions to $x^{\sin y}=y^{\sin x}$, in agreement with the graph in Ákos Somogyi's answer.
