(a) $x=-2 \sin (3 t+\pi / 3)$ $=2 \cos (3 t+\pi / 3+\pi / 2)$ $=2 \cos (3 t+5 \pi / 6)$
On comparing the above equation with the standard equation, $x=A \cos (\omega t+\Phi)$,
Amplitude will be $A=2 \mathrm{~cm}$ (radius of the circle)
Angular velocity will be $\omega=3 \mathrm{rad} / \mathrm{s}$
Phase angle will be $\Phi=5 \pi / 6=150^{\circ}$
(b) $x=\cos (\pi / 6-t)$ $=\cos (t-\pi / 6)$
On Comparing this equation with $A \cos (\omega t+\Phi)$, we get,
Phase angle as $\Phi=-\pi / 6=-30^{\circ}$
Amplitude as $A=1 \mathrm{~cm}$ (radius of the circle)
Angular velocity as $\omega=1 \mathrm{rad} / \mathrm{s}$