Consider the $AR(1)$ process given by $(1-0.6B)(X_{t} - 3) = a_{t}$ where $a_{t} \sim WN(0,1)$ and $B$ is the backshift operator ($X_{t}B = X_{t-1}$).
We can rewrite the process in the more convenient form: $$X_{t} = 0.6 X_{t-1} + 1.2 + a_{t}$$
By way of recursively substitution, we can obtain an $MA(\infty)$ representation, as by $X_{t} = \displaystyle\sum_{j=0}^{\infty} 0.6^{j} a_{t-j}$, from which it follows that $\gamma(0) = \frac{1}{1-0.6^{2}} = 1.5625$, where $\gamma(k)$ is the autocovariance funtion at lag $k$.
I was interested in determining $\gamma(1)$ and $\gamma(2)$, as well. To do this, I multiplied the above equation by $X_{t-1}$ and $X_{t-2}$ respectively and took the expectation of both sides. This gave the following relations:
$$\gamma(1) = 0.6\gamma(0) + 1.2 \text{ and } \gamma(2)=0.6\gamma(1) + 1.2$$
Everything I've done is pretty straightforward. However, if we multiply the above equation by $X_{t}$ on both sides, taking expectation gives us another expression $\gamma(0) = 0.6\gamma(1) + 1.2 + \sigma^{2}$ which disagrees with the previous determination.
Why do these two determinations of $\gamma(0)$ disagree? What have I done wrong?