Let $X$ be a truncated normal random variable with original (pre-truncated) mean $\mu$, variance $\sigma^2$, and left and right truncation points $a$ and $b$ respectively. It's known that the expected value of $X$ is
$$ E(X) = \mu + \frac{\phi\left(\frac{a-\mu}{\sigma}\right)-\phi\left(\frac{b-\mu}{\sigma}\right)}{\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)} $$
where $\phi$ is the pdf of the standard normal and $\Phi$ is the cdf of the standard normal.
But: if I know what $E(X)$ is equal to, as well as the values of $\sigma, a$, and $b$, how do I invert this expression for $\mu$? Is that even possible?
Intuition says that there should be a unique solution, but I can't quite seem to calculate what it is. Surely the answer will involve the inverse pdf and cdf of the standard normal, but I keep getting stuck. Any pointers?