About Root mean square voltage
Even though its intuitive why it's zero. We can just integrate to find that result.

Solution:

$$\int_0^Tcos(2\omega t+\phi)$$
$$=\int_0^Tcos(2.\frac{2\pi}{T} t+\phi)$$
$$=\int_0^Tcos(\frac{4\pi}{T} t+\phi)$$
$$=\left[\frac{sin(\frac{4\pi}{T}t+\phi)}{4\pi/T}\right]^T_0$$
$$=\frac{sin(4\pi+\phi)-sin\phi}{\frac{4\pi}{T}}$$
$$=\frac{sin4\pi.cos\phi+sin\phi.cos4\pi-sin\phi}{\frac{4\pi}{T}}$$
$$= \frac{0.cos\phi+sin\phi.1-sin\phi}{\frac{4\pi}{T}}$$
$$=  \frac{sin\phi-sin\phi}{\frac{4\pi}{T}} $$
$$=0$$