I came to this frequency spectrum for a signal $h_1[n]$:
\begin{align*} H_1(e^{j\theta}) &= \sum_{n=-\infty}^{\infty} h_1[n] \cdot e^{-j\theta} \\ &= \sum_{n=-\infty}^{\infty} (0.3 \cdot \delta[n] + 0.6 \cdot \delta[n-1] + 0.3 \cdot \delta[n-2]) \cdot e^{-j\theta} \\ &= \ldots \\ &= 0.6 e^{-j\theta} \cdot (1 + cos(\theta) ) \end{align*}
I also have the magnitude spectrum:
\begin{align*} |H_1(e^{j\theta})| &= |0.6 e^{-j\theta} \cdot (1 + cos(\theta) )| \\ &= 0.6 \cdot \underbrace{|e^{j\theta}|}_\text{$=1$} \cdot |(1 + cos(\theta))| \\ &= 0.6 \cdot |(1 + cos(\theta))| \geq 0 && \forall \theta \in \mathbb{R} \end{align*}
And now I am supposed to give the phase spectrum of this. My problem is that I don't know how this $arg(\cdot)$ function works here. For
\begin{align*} arg( 0.6 e^{-i \theta} (1 + cos(\theta) ) ) &= -2 \theta - arg( 0.6 e^{-i \theta} (1 + cos(\theta) ) ) \end{align*}
Could somebody explain this to me?
E.g. why is for:
\begin{align*} H_1(e^{j\theta}) &= 0.5 + 0.86 \cdot cos(\theta) \end{align*}
The $arg(\cdot)$:
\begin{align*} arg(H_1(e^{j\theta})) &= - 2 \theta - arg(0.5 + 0.86 \cdot cos(\theta)) \end{align*}
You correctly computed the complex frequency response as
$$H(e^{j\theta})=0.6(1+\cos(\theta))e^{-j\theta}\tag{1}$$
This is exactly in the form
$$|H(e^{j\theta})|e^{j\phi(\theta)}\tag{2}$$
where $\phi(\theta)$ is the desired phase function, because, as you've noted yourself,
$$0.6(1+\cos(\theta))\ge 0$$
so it qualifies as the magnitude of $H(e^{j\theta})$. Comparing (1) and (2) gives
$$\phi(\theta)=-\theta\tag{3}$$
so the phase is linear, which is also obvious from the symmetry of the sequence $h[n]$.