Differentiating the mean-field free energy and imposing stationarity gives
$$
0=Jqm-Jq\frac{\sinh(\beta Jqm)}{1+\cosh(\beta Jqm)}.
$$
The <hyperbolic-function identity> $\sinh x/(1+\cosh x)=\tanh(x/2)$ turns this into the <self-consistency equation>
$$
\boxed{m=\tanh\!\left(\frac{Jqm}{2T}\right)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 / 1 / b / ii / Solution

Using the field derivative while imposing the <self-consistency equation> gives
$$
\boxed{m=\frac{2\kappa\sinh[\beta(Jqm+B)]}
{1+2\kappa\cosh[\beta(Jqm+B)]}}.
$$
At $B=0$, $m=0$ is always a solution. Linearizing the right-hand side for small $m$ gives slope
$$
\frac{2\kappa\beta Jq}{1+2\kappa}.
$$
A continuous <phase transition> occurs when this slope crosses one, so, with $k_B=1$ and $\kappa_c=e^{-g/T_c}$,
$$
\boxed{1=\frac{2e^{-g/T_c}}{1+2e^{-g/T_c}}\frac{Jq}{T_c}}.
$$
On the side where the slope exceeds one, the nonzero solutions lower the mean-field free energy and the trivial solution is unstable. The equation has a physical continuous-transition solution only in the corresponding parameter range of this spin-one model.

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 303 / 1 / a / iii / Solution

A self-consistency equation requires the average field assumed in a <mean-field approximation> to equal the expectation value calculated in that field.

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