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I am concerened I may have oversimplified my solution to this question.

My solution:

Let $F(x,y,z)=x-e^y\sin(z)$

By the implicit function theorem: $\displaystyle\frac{\partial z}{\partial x}=-\displaystyle\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}$

$$\frac{\partial F}{\partial z}=-e^y\cos(z)$$

As $e^y\neq0$ by the implicit function theorem $\cos(z)\neq0$

Michael Hardy
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usainlightning
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    I changed $sin(z)$ and $cos(z)$ to $\sin(z)$ and $\cos(z)$. Writing \sin(z) instead of sin(z) not only prevents $\sin$ from being italicized, but also provides proper spacing in expressions like $x\sin y$. It is standard usage. – Michael Hardy Jun 23 '14 at 16:31
  • You can explicitly write a local solution $z = \sin^{-1}(x e^{-y})$. So you need to add the condition that $x e^{-y} \in [-1,1]$. (By "solution", you did mean "solution that is continuous in a neighborhood", didn't you?) – Stephen Montgomery-Smith Jun 23 '14 at 16:39
  • I think that $x,y,z$ here are assumed to be real. If this is the case then only when$x \exp{-y}le 1$ can we express $z=\arcsin(x \exp{-y})$ – mike Jun 23 '14 at 16:39

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The implicit function theorem says that you can define $z$ as a function of $x$ and $y$ if and only if $\partial f/\partial z=-e^y\cos z\neq 0$, that is, $z/\pi\notin \Bbb Z$.

ajotatxe
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