= Solution
Expanding $R\sin(x+\phi)$ and equating coefficients gives $R\sin\phi=A$, $R\cos\phi=B$. Thus
$$
\boxed{R=\sqrt{A^2+B^2},\qquad \phi=\operatorname{atan2}(A,B)\pmod{2\pi}.}
$$
The two-argument <arctangent> fixes the quadrant; if $R=0$, the phase is arbitrary.
For the coupled <ordinary differential equations>, set $z=fg$. The first equation becomes $\dot z=-3z$, and the initial values give $z=2e^{-3t}$. The second then reduces to $f''+3f=12e^{-3t}$. Its <initial value problem> has solution
$$
\boxed{f(t)=e^{-3t}+\cos(\sqrt3t)+\sqrt3\sin(\sqrt3t)=e^{-3t}+2\cos\left(\sqrt3t-\frac\pi3\right).}
$$
Where $f\ne0$, reconstruct $g(t)=2e^{-3t}/f(t)$. The scalar function $f$ is smooth for every real $t$, but the original coupled system cannot have a finite $g$ at its first zero, about $t=1.5146$. The requested sketch up to $t=3$ can display the analytic continuation of $f$, with that distinction marked. Initially $f=2$, $f'=0$ and $f''=6$; it first rises, reaches a maximum, then decreases through zero and becomes negative within the plotted interval.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2016-coupled-ode-determinant.png]
{title=The scalar solution and matrix determinant, with the first regular rotation time and the later pole in the reconstructed variable}
For the <determinant> of $M$, separate its upper-left block from the constant last diagonal entry. Using $fg=2e^{-3t}$ gives
$$
\det M=\frac34+\frac{f(2-g)}{16}=\frac34+\frac{\cos(\sqrt3t)+\sqrt3\sin(\sqrt3t)}8.
$$
Thus $\det M=1$ precisely when $2\cos(\sqrt3t-\pi/3)=2$. The smallest positive solution is
$$
\boxed{t_*=\frac\pi{3\sqrt3}.}
$$
It occurs before the zero of $f$, so $g$ and $M$ are finite there. Put $f_*=2+e^{-\pi/\sqrt3}$. At this time the upper block is
$$
\begin{pmatrix}f_*/(2a)&-\sqrt3/2\\\sqrt3/2&a/(2f_*)\end{pmatrix}.
$$
For a <rotation matrix>, each column must have unit norm, giving $a^2=f_*^2$. Both signs are allowed by the stated assumption that $a$ is a constant:
$$
\boxed{\begin{array}{c|c|c}
a&\text{axis}&\text{angle about positive }z\\\hline
2+e^{-\pi/\sqrt3}&z&\pi/3\\
-(2+e^{-\pi/\sqrt3})&z&2\pi/3
\end{array}}
$$
Both have <determinant> $1$ and preserve lengths. If a positive $a$ is intended, only the first row remains; positivity is not stated in the question.
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