Solution
= Solution
The half-power amplitude condition for the <RC low-pass filter> gives $1+(\omega RC)^2=2$. Since <angular frequency> is positive,
$$
\boxed{\omega_c=\frac1{RC}.}
$$
At this <cutoff angular frequency>, the output amplitude is smaller by $1/\sqrt2$ and the <phase shift> is $-\pi/4$.