What will be the domain of $ \lg(\sin x \cos y) $? I know that $ \lg x>0$, which $ \implies$ $(\sin x \cos y)>0$ and I assume that it will be a multivariable function because there are two variables, $(x,y)$ and the domain of $\sin x$ and $ \cos x$ will be between $-1$ and $1$, how can I continue it? Are my toughts correct? How can I plot the domain?
 A: You are mixing the concepts of range and domain. The domain is the set of values for the independent variables for wich the function is defined, in this case is a subset of $\mathbb R^2$. The range is the set of values the function reaches. In this case, a subset of $\mathbb R$.
$\ln g(x)$ is defined for values $g(x)\gt0$ and it's range can even be $\mathbb R$ or more limited depending on the range of $g(x)$
Domain (corrected)
$\sin x\cos y\gt0\implies (\sin x\gt0\land\cos y\gt0)\lor(\sin x\lt0\land\cos y\lt0)$ So is, both positive or both negative. Meaning that (considering that $0\in\mathbb N$)
$D=\{(x,y)\in\mathbb R^2\vert\\
\vert\,(2\pi k_1\lt x\lt\pi(2k_1+1),k_1\in\mathbb N;\land\;-\pi/2+2\pi k_2\lt y\lt\pi/2+2\pi k_2,k_2\in\mathbb N)\lor\\
\lor(\pi(k_3+1)\lt x\lt2\pi(k_3+1),k_3\in\mathbb N\;\land\;\pi/2+2\pi k_4\lt y\lt3\pi/2+2\pi k_4,k_4\in\mathbb N\}$
The domain resembles a chessboard with the, say, black squares in the domain and the white ones excluded of it.
Range
$\log g$ is a monotonically increasing function, so, in order to determine the range, we only need to check the  minimum value for its argument and the maximum one.
$\sin 0=0\implies\ln(\sin0\cos y)=-\infty$
The maximum for $\sin x$ is reached for $\pi/2$, $\sin(\pi/2)=1$ and the one for $\cos y$ for $y=0$. Then the maximum for the logarithm is $\ln(\sin(\pi/2)\cos 0)=\ln 1=0$
$R=\{z\in\mathbb R\,\vert\,z\lt0\}$
