question related to something in karlin and taylor stochastic processes one text
The mean, i.e., the expected value of $X_n$ is

$$E\{X_n\}=E\{A\cos(\omega n)+B\sin(\omega n)\}\tag{1}$$

Since the expectation operator is linear, $(1)$ is equivalent to

$$E\{X_n\}=E\{A\cos(\omega n)\}+E\{B\sin(\omega n)\}\tag{2}$$

and since $\cos(\omega n)$ and $\sin(\omega n)$ are deterministic you finally get

$$E\{X_n\}=E\{A\}\cos(\omega n)+E\{B\}\sin(\omega n)\tag{3}$$

which equals zero if both $E\{A\}$ and $E\{B\}$ are zero. If at least one of the two random variables $A$ and $B$ has a non-zero mean, then $(3)$ will generally depend on $n$ (except for very specific choices of $\omega$), and, consequently, the mean of $X_n$ will also depend on $n$, i.e., $X_n$ cannot be stationary. 

As a side note, a constant mean does of course generally not imply stationarity of $X_n$. For the given process, however, if $A$ and $B$ are identically distributed and $E\{A\}=E\{B\}=0$, and under the additional assumption that $E\{AB\}=0$ holds, $X_n$ is a wide-sense stationary (WSS) process.