Formula is that:
$$V=V_{m}\cos(\omega t+\phi)$$
And question says:
Question) $V=300\cos(120\pi t + 30^{\circ} )$ What is the magnitude of V at t=2.778mss?
I didn't understand that in the solution we didn't directly use value of the t. Because the solution is:
$$\omega t = \frac{2\times\pi\times2.778\times 10^{-3} }{16.667\times10^{-3}}$$
$$v(2.778ms)= 300\cos(60^{\circ}+30^{\circ})= 0\,V $$
At $t = 2.778 ms$,
$V(t) = 300 \cos( 120 \times \pi \times 2.778 \times 10^{-3} + 30^\circ)$
Since $30^\circ = \pi/6$, you can write:
$V(t) = 300 \cos( 120 \times \pi \times 2.778 \times 10^{-3} + \pi/6)$
$V(t) \approx 300 \cos(\pi/3 + \pi/6) = 300 \cos(\pi/2) = 0$
There is some numerical approximation here. $120 \times 2.778 \times 10^{-3}$ is approximately 1/3, but not exactly. You can of course plug in the exact value of $t$ to get the exact answer $-0.02513$.
It is not a very well framed question or solution.