Autocorrelation of the product of deterministic and random signal
As correctly pointed out in the comments, in general the process $Y(t)=x(t)M(t)$ is not wide-sense stationary (WSS), i.e. its autocorrelation function depends not only on the time difference parameter $\tau$, but also on the absolute time $t$:

$$R_Y(\tau,t)=E[Y(t+\tau)Y^*(t)]=E[x(t+\tau)M(t+\tau)x^*(t)M^*(t)]=\\
=x(t+\tau)x^*(t)E[M(t+\tau)M^*(t)]=x(t+\tau)x^*(t)R_M(\tau)\tag{1}$$

In your case you just need to evaluate (1) with the given function $x(t)$. However, as expected you will end up with a function of two variables because $Y(t)$ is not WSS.

There are a few special cases in which the resulting process is indeed WSS or can be easily made WSS. The first case is modulation by a complex exponential:

$$x(t)=e^{j\omega_ct}$$

in which case

$$x(t+\tau)x^*(t)=e^{j\omega_c(t+\tau)}e^{-j\omega_ct}=e^{j\omega_c\tau}$$

only depends on $\tau$ and not on $t$. Another case of interest is the case where $x(t)$ is $T$-periodic. In this case the process $x(t)M(t)$ can be made WSS by introducing a random phase epoch $\Theta$ with uniform distribution in the interval $[0,T]$ which is independent of $M(t)$:

$$Y(t)=x(t+\Theta)M(t)$$

The autocorrelation function of $Y(t)$ is then given by

$$R_Y(\tau)=R_M(\tau)\frac{1}{T}\int_0^Tx(\alpha+\tau)x^*(\alpha)d\alpha$$