Help me understand the phase of an FIR filter I'm going over an exercise of designing an simple FIR filter, I have the solution in front of me, but I struggle to understand some parts of it.

I'm asked to have an FIR filter with only 3 coefficients.

I have reached this form:
$$ H(\theta ) = (2\alpha_0 \cos(\theta)+\alpha_1)e^{-j\theta} $$
where $\theta$ is the frequency variable (DTFT).

The magnitude is easy:
$$ \left | H(\theta ) \right |=\left | 2\alpha_0 \cos(\theta)+\alpha_1 \right | $$ 

But I struggle with the phase, it seems easy: $\phi(H(\theta))=-\theta$ , but I guess this is wrong ? why ?

The solution that is given is:

$$\phi(H(\theta)) = -\theta +\beta$$
$$\beta = 0 \text{ for } 2\alpha_0 \cos(\theta) + \alpha_1 &gt; 0$$

$$\beta = \pi \text{ for } 2\alpha_0 \cos(\theta) + \alpha_1 &lt; 0$$

EDIT:

I think I just answers figured it out, if that expression is negative we can write it as positive with a phase of π. What happens if $\alpha_1$ is larger than $4\alpha_0$ thus the expression is always positive ? does $\beta =0$ ?
Your answer does not take into account the possible sign of the magnitude term in the polar description of the frequency response $H(\theta) = A(\theta)e^{j\phi(\theta)}$. I guess you mix this definition with the more usual definiton of  $H(\theta) = |A(\theta)|e^{j\phi(\theta)}$ where the magnitude will always be positive and therefore phase will be what you predicted, however for your example $A(\theta)$ may assume a negative value and you shall reflect this as an addition of $\pi$ into the phase where $e^{j(\theta + \pi)} = -e^{j\theta} $