= Solution
Let $\mathbf u_\theta=(\cos\theta,\sin\theta)^{\mathsf T}$ be the transmission direction. An ideal linear <polarizer> projects the incident <electric field> onto this direction, so its <Jones matrix> is
$$
\boxed{P(\theta)=\mathbf u_\theta\mathbf u_\theta^{\mathsf T}
=\begin{pmatrix}\cos^2\theta&\cos\theta\sin\theta\\
\cos\theta\sin\theta&\sin^2\theta\end{pmatrix}}.
$$
Since $\mathbf u_\beta^{\mathsf T}\mathbf u_\alpha=\cos(\alpha-\beta)$, the <amplitude> transmitted by a sequence of <polarizers> is the product of the projections between successive axes. Here the four directions are $0,\omega t,-2\omega t,0$. After the first <polarizer> the <unpolarized light> input has intensity $I/2$ and is in <linear polarization> along $x$. The <transmission through a sequence of linear polarizers> therefore follows from successive applications of <Malus's law>
$$
\boxed{I_{\rm out}(t)=\frac I2\cos^2(\omega t)\cos^2(3\omega t)\cos^2(2\omega t)}.
$$
The relative angle between the rotating <polarizers> is $3\omega t$, not $\omega t$. The first factor $1/2$ is an <ensemble average> over incident polarization: <unpolarized light> cannot be represented by a single deterministic <Jones vector>. Equivalently, its <coherency matrix> is $(I/2)I_2$; the total <matrix> $P(0)P(-2\omega t)P(\omega t)P(0)$ gives the same output intensity by the <trace> rule. The quasistatic optical-element approximation assumes the mechanical rotation is slow compared with the optical oscillation.
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