Calculating the system output using frequency response Given an input signal $$x(n)=\cos(6\pi n +\frac{\pi}{6})$$ and system $$y(n)=0.5x(n)-0.1x(n-1)$$. In this case, the coefficients of the difference equation are $a_0=1$, $b_0=0.5$, and $b_1=$. The frequency response to this system is $$H(\omega)= b_0 + b_1e^{-j\omega}= 0.5 - 0.1 \cos(\omega)+0.1 j \sin(\omega)$$  

Since the input $x(n)$ is a sinusoid of the form $$x(n)=A \cos(\omega n +\theta_0)$$, we can express the system's output as  

$$y(n)=A|H(\omega_0)|\cos\left(\omega_0 n+\theta_0+angle(H(\omega_0))\right)$$

I can see that $A=1$, $\omega_0=6\pi$, and $\theta_0=\frac{\pi}{6}$ in this formula.  Hence, I have deduced that  

$$|H(\omega_0)|= |0.5 - 0.1 \cos(6\pi)+0.1j \sin(6\pi)| = 0.4$$

Furthermore, since $H(\omega_0) = 0.4$ (a positive real number), I conjecture that $$angle(H(\omega_0))=0$$

Have I made the correct assumptions/conclusions?
Yes, you are correct.  The system input is rather strange, though, since $cos(6\pi n + \frac{\pi}{6})$ is equal to $cos(\frac{\pi}{6})$.