Given $\mathbb Q$ and $X_t$ is $\mathbb Q$-Brownian, find $\frac{d\mathbb Q}{d\mathbb P}$ / Uniqueness of Brownian or Radon-Nikodym derivative
As you say (how did $\sin t$ in the initial problem become $\cos t$?), $\mathbb{Q}$ is a measure under which $W_t$ becomes $\tilde{W}_t - \sin t$, where $\tilde{W}_t$ is a $\mathbb{Q}$-Brownian motion. So Girsanov's theorem would say

$$
L_t = E[\frac{ d \mathbb{Q} }{ d \mathbb{P} }|\mathcal{F}_t] = e^{ - \int_0^t \sin s dW_s - \frac{1}{2} \int_0^t \sin^2 s ds}
$$

is one such measure.

As for uniqueness, if we take $\Omega$ to be the canonical path space $C[0, T]$ with $\mathcal{F}$ generated by the cylinder sets. Then the Wiener measure would be unique, simply by the fact that the Brownian law says any two such measures must agree on the cylinder sets and therefore all of $\mathcal{F}$.

In general, the Girsanov $\mathbb{Q}$ is unique up to the Brownian filtration $\mathcal{F}_{W} \subset \mathcal{F}$, since it is constructed $W$-pathwise.