Evaluate phase of $X(\omega)$ without computing $X(\omega)$ $$x(n) = \{ -1, 0, 1, 2, 1, 0, 1, 2, 1, 0, -1 \}$$
Let $X(\omega)$ be the DTFT of $x(n)$. I need to find the phase of $X (\omega)$ without computing $X(\omega)$. I notice that $x(n)$ can be a type I linear phase filter, but not sure if that is used in solving the problem.
hint: x(n) has symmetric coefficients which as a filter is a type 1 linear phase filter as the OP has stated. Such a filter will have a phase slope that is dependent on the number of taps only.