 Solution

Expanding $R\sin (x+\varphi )$ and equating coefficients gives $R\sin \varphi =A$, $R\cos \varphi =B$. ThusThe 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 $f^{\prime \prime }+3f=12e^{-3t}$. Its initial value problem has solutionWhere $f\ne 0$, 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 $f=2$, $f^{\prime }=0$ and $f^{\prime \prime }=6$; it first rises, reaches a maximum, then decreases through zero and becomes negative within the plotted interval.

For the determinant of $M$, separate its upper-left block from the constant last diagonal entry. Using $fg=2e^{-3t}$ givesThus $\det M=1$ precisely when $2\cos (\sqrt{3}t-\pi /3)=2$. The smallest positive solution isIt occurs before the zero of $f$, so $g$ and $M$ are finite there. Put $f_{\ast }=2+e^{-\pi /\sqrt{3}}$. At this time the upper block isFor a rotation matrix, each column must have unit norm, giving $a^2=f_{\ast }^2$. Both signs are allowed by the stated assumption that $a$ is a constant: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.