= Solution
Set $x=T^4/T_0^4-1$. Since $\sigma\gamma T_0^4=Q/8$, the <steady state> <energy balance> is $2(1-a_0)=1+x$. Substitution gives
$$
\boxed{\tanh(\mu x)=2x.}
$$
The two sides have slopes $\mu$ and $2$ at the origin. For $x>0$, $\tanh(\mu x)$ is strictly <concave>, its initial slope is $\mu$, and it remains below $1$. Consequently no positive root exists for $\mu\leq2$, while for $\mu>2$ there is exactly one positive root $x_+\in(0,1/2)$. Oddness supplies the corresponding negative root. At $\mu=2$ the only root is still zero. Hence \b[multiple steady temperatures occur precisely when $\mu>\mu_c=2$].
The largest root corresponds monotonically to the largest positive absolute <temperature>:
$$
\boxed{T_+=T_0(1+x_+)^{1/4},\qquad a_+=\frac{1-x_+}{2}.}
$$
For large $\mu$, the positive root tends to $1/2$, so $a_+\simeq1/4$ and $T_+/T_0\simeq(3/2)^{1/4}$. The negative branch has larger <albedo> and lower <temperature>.
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