= Solution
The operator is \b[well-defined in every dimension]: $|\cos w|\leq1$ gives
$$
|\langle A(w,u),v\rangle|\leq\|\nabla u\|_2\|\nabla v\|_2\leq\|u\|_{H^1}\|v\|_{H^1}.
$$
For fixed $w$, its monotonicity condition is
$$
\langle A(w,u)-A(w,z),u-z\rangle=\int_\Omega\cos(w)|\nabla(u-z)|^2\geq0.
$$
Thus $\cos w\geq0$ <almost everywhere> is sufficient. It is also necessary, by the <positivity criterion for a scalar gradient form>. Indeed, if the bounded coefficient $c=\cos w$ is negative on a set of positive measure, choose a <Lebesgue point> $x_0$ with $c(x_0)<0$ and a nonconstant compactly supported smooth $\psi$. Set $z_\varepsilon(x)=\varepsilon\psi((x-x_0)/\varepsilon)$, supported inside $\Omega$. Then
$$
\varepsilon^{-n}\int_\Omega c(x)|\nabla z_\varepsilon(x)|^2\,dx\longrightarrow c(x_0)\int|\nabla\psi|^2<0,
$$
contradicting monotonicity. Therefore $\boxed{A(w,\cdot)\text{ is monotone exactly when }\cos w\geq0\text{ almost everywhere}}$. In particular it is not monotone for every $w$: the admissible constant $w\equiv\pi$ makes it the negative Dirichlet form.
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