How to derive the sample size $n$ to achieve $P(\text{type 2 error}) = \beta$ for a two-tailed test? Consider testing $H_0 : \mu = \mu_0$ vs $H_a : \mu \neq \mu_0$ at level $\alpha$. We then wish to control for the type II error, that is have $P(\text{type II error} \ | \ \mu = \mu_1) = \beta$ for a predetermined $\beta$
Assume $\bar{X}$ is approximately normal with mean $\mu$ and standard deviation $\sigma / \sqrt{n}$. Base a hypothesis test on the standardized variable $Z = \frac{\bar{X}-\mu_0}{\sigma/\sqrt{n}}$, where $\bar{X} = \frac{1}{n} \sum_{i=1}^n X_i$
Then the power function is 
$\gamma(\mu) = P(\text{reject} \ H_0 \ | \ \mu) 
= P(\frac{\bar{X}-\mu_0}{\sigma/\sqrt{n}} \geq -z_{\alpha/2} \ \text{or} \ \frac{\bar{X}-\mu_0}{\sigma/\sqrt{n}} \leq z_{\alpha/2} \ | \ \mu)$ 
$= P(\frac{\bar{X}-\mu_0}{\sigma/\sqrt{n}} \geq -z_{\alpha/2} \ | \ \mu) + P(\frac{\bar{X}-\mu_0}{\sigma/\sqrt{n}} \leq z_{\alpha/2} \ | \ \mu) = 1-\Phi(z_{\alpha/2} + \frac{\mu_0-\mu}{\sigma/\sqrt{n}}) + \Phi(-z_{\alpha/2} + \frac{\mu_0-\mu}{\sigma/\sqrt{n}})$
where we have solved the inequalities for $\bar{X}$ and restandardized with $\mu_1$.
Then, for $\mu_1 \neq \mu_0$,
$P(\text{type II error} \ | \ \mu=\mu_1) = P(\text{don't reject} \ H_0 \ | \ \mu=\mu_1)$
$=1 - P(\text{reject} \ H_0 \ | \ \mu=\mu_1) = 1 - (1-\Phi(z_{\alpha/2} + \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}}) + \Phi(-z_{\alpha/2} + \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}}))$
$= \Phi(z_{\alpha/2} + \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}}) + \Phi(-z_{\alpha/2} + \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}})$
$ = \beta$
How to solve this for $n$? The answer is supposed to be $n = \bigg(\frac{\sigma (z_{\alpha/2} + z_{\beta})}{\mu_0 - \mu_1}\bigg)^2$ (see Devore & Berk, Modern Mathematical Statistics with Applications, 2012: page 441)
Sidenote: The way I solved it in the one-tailed case was simply by doing
$\Phi(z_{\alpha/2} + \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}}) = \beta$
$\Phi^{-1}\Phi(z_{\alpha/2} + \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}}) = \Phi^{-1}\beta = -z_\beta$, etc
 A: The approximated solution can be derived as below.
\begin{align*}
& {\qquad}
1-\beta 
= 
\gamma(\mu) \\  
& {\qquad}
= 
1 + 
\Phi
\left(
k-z_{\alpha/2}
\right)
-
\Phi
\left(
k+z_{\alpha/2}
\right),
\quad \mbox{where} \quad 
k := \frac{\mu_0-\mu}{\sigma/\sqrt{n}} \\
& {\qquad} = 
P(Z \ge z_{\alpha/2}-|k|) + P(Z \ge z_{\alpha/2}+|k|) \\
\Rightarrow & {\qquad} 
1-\beta \approx P(Z \ge z_{\alpha/2}-|k|),
\quad \mbox{assuming} \quad P(Z \ge z_{\alpha/2}+|k|) \approx 0 \\
\iff & {\qquad} 
z_{1-\beta} \approx z_{\alpha/2}-|k| \\
\iff & {\quad}
-z_{\beta} \approx z_{\alpha/2}-|k| \\
\iff & {\quad} 
|k| \approx  z_{\alpha/2}+z_{\beta},
\end{align*}
this gives 
$$
n \approx
\left[
\frac{\sigma(z_{\beta} + z_{\alpha/2})}
{\mu_0-\mu}
\right]^2,
$$
as desired.
A: It is helpful to perform the calculation for a specific set of parameters.  Let's choose $\mu_0 = 0$ and $\mu_1 = 1$, with $\sigma = 1$.  Moreover, let $\alpha = 0.05$.  We wish to find $n$ such that the test has power $1 - \beta = 0.9$.  For the two-sided hypothesis $$H_0 : \mu = \mu_0 \quad \text{vs.} \quad H_a : \mu \ne \mu_0$$ the test statistic $$Z = \frac{\bar X - \mu_0}{\sigma/\sqrt{n}} = \bar X \sqrt{n}$$ will result in rejection of $H_0$ in favor of $H_a$ if $$|Z| > z_{\alpha/2}^* \approx 1.96,$$ where I have used the asterisk to indicate the quantile is the upper, rather than lower quantile.  Therefore, failure to reject $H_0$ when $H_1 : \mu = \mu_1$ is true, is $$\beta = \Pr[|Z| \le z_{\alpha/2}^* \mid \mu = \mu_1] = \Pr[-1.96 \le \bar X\sqrt{n} \le 1.96 \mid \mu = \mu_1] = 0.1.$$  Since $$\bar X \sqrt{n} \mid H_1 \sim \operatorname{Normal}(\mu = \sqrt{n}, \sigma = 1),$$ it follows that we want to find the smallest positive integer $n$ such that $$\Pr[-1.96 - \sqrt{n} \le Z \le 1.96 - \sqrt{n}] \le 0.1$$ in order for the test to have Type II error control at level $\beta$.  This is where you get stuck, because changing $n$ changes both endpoints of the inequality, so finding an inverse function is not tractable.  But if we observe that the lower endpoint is already very far to the left--that is to say, $\Pr[Z < -1.96 - \sqrt{n}] \approx 0$ for any reasonable choice of positive integer $n$--then, we can ignore this condition and invert $$\Pr[Z \le 1.96 - \sqrt{n}] \le 0.1$$ to get $$1.96 - \sqrt{n} \le z_\beta = z_{0.1} \approx -1.282.$$  Consequently $$n \ge 10.5106,$$ which rounded up yields $n = 11$.  The ignored tail probability is $$\Pr[Z \le -1.96 - \sqrt{10.5106}] = \Phi(-5.202) \approx 9.86 \times 10^{-8}.$$  Using a computer, we can numerically solve for the precise $n$ that meets the two-sided inequality:  $$n \approx 10.50741940969075474768$$ to 20 decimal places.  But this is effectively unnecessary since this $n$ is always less than the $n$ obtained by ignoring the smaller tail, since doing so means that a larger $n$ must be chosen to ensure Type II error control.
Now that we understand this numeric example, it is not too hard to revisit the general case, being careful to correctly identify which tail probability is smaller.
