Linear-phase lowpass filters: frequency responses alignment
Both methods are equivalent. Look at it like this: first write down the frequency responses (as already indicated in your question)

$$H_1(e^{j\omega})=1+2e^{-j\omega}+e^{-2j\omega}\\
H_2(e^{j\omega})=1+e^{-j\omega}$$

Now want to find $\omega_0$ such that

$$H_1(e^{j\omega_0})=H_2(e^{j\omega_0})$$

which gives

$$1+2e^{-j\omega_0}+e^{-2j\omega_0}=1+e^{-j\omega_0}$$

or, equivalently,

$$e^{-j\omega_0}(1+e^{-j\omega_0})=0\tag{1}$$

Equation (1) can only be satisfied if the term in parentheses vanishes. So you get

$$e^{-j\omega_0}=-1\Longrightarrow\omega_0=\pi$$

And since $\omega_0=2\pi f_0/f_s$ (where $f_s$ is the sampling frequency) you finally get

$$f_0=f_s/2$$