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I have a random process defined as

$$X(t) = A sin(omega t + phi)$$

where A and $omega$ are independent. $phi$ is distributed $U[0,2pi]$.

I would like to find $E[X(t)]$.

I believe the answer is $0$ because the multiplication of 2 expectation of independent random variables is the product of the expectation of each. i.e. $E[X(t)] = E[A] E[sin(omega t + phi)]$.

Since $phi$ is distributed uniformly over a cycle, the mean of the $sin()$ term must be zero thereby making the whole quantity 0.

Is this correct?

Yes, if $(A,omega,phi)$ is mutually independent.

Indeed, if $(A,omega,phi)$ is mutually independent, then $A$ is independent of $(omega,phi)$ so you can write $mathbb E[X(t)]=mathbb E[A]mathbb E[sin(omega t+phi)]$.

Moreover, $omega$ is independent of $phi$ so
$$
mathbb E[sin(omega t+phi)]=mathbb E[sin(omega t)cos(phi)]+mathbb E[sin(phi)cos(omega t)]=mathbb E[sin(omega t)]mathbb E[cos(phi)]+mathbb E[sin(phi)]mathbb E[cos(omega t)]=0
$$

I used the formula $sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$ because it requires the least knowledge in probability theory. If you are more familiar with the probability theory, here is another way to conclude: since $omega$ and $phi$ are independent, we have
$$
mathbb E[sin(omega t+phi)]=mathbb E[f(omega t)],
$$
where $f:mathbb Rtomathbb R$ is defined for all $xinmathbb R$ by $f(x)=mathbb E[sin(x+phi)]=0$.

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&bbox[10px,#ffd]int_0^2piAsinparsomega t + phi,ddphi over 2pi =
left.phantomLarge A
-,A over 2pi,cosparsomega t + phi,rightvert_ phi = 0^ phi = 2pi
\[5mm] = &
-,A over 2pi,cosparsomega t + 2pi + A over 2pi,cosparsomega t + 0 =
bbx0
endalign

$dscos$ is a periodic function of period $ds2pi$.