I thought I'd post an answer since I figured this one out an it had a few views:
$$H(s)=\frac{10(s+300)}{s^2+20s+50000}$$
It should be immediately obvious the poles are complex, but we only need to know them to determine \$h(t)\$ later.
The input has a frequency \$\omega=2\pi30\$ in \$x(t)=2\cos(2{\pi}30t + 0.7)u(t)\$
Substituting for \$s=j\omega\$ we get
$$H(j\omega)=\frac{10j\omega+3000}{20j\omega+50000-\omega^2}$$
The magnitude and phase of this response at a chosen frequency affects the forced response's magnitude and phase. ie for a sinusoidal input:
$$u(t)=M_i\cos({\omega}t+{\phi}_i)\tag{1}$$
$$y(t)=M_o(\omega)\cos({\omega}t+{\phi}_o(\omega))\tag{2}\label{2}$$
Where the subscripts \$i, o\$ denote the input and output magnitudes and phases.
Substituting \$\omega=2{\pi}30\$ gives
$$\frac{3000+j600\pi}{50000-3600\pi+j1200\pi}\tag{3}$$
$$=0.09114{\angle}0.4639=M_o(\omega)\angle{\phi}_o(\omega)\tag{4}\label{4}$$
With \$\ref{4}\$ substituted into \$\ref{2}\$
The sinusoidal steady-state response is
$$x(t)=0.1823\cos(2{\pi}30t + 1.1622)u(t)$$
Which is the desired result.
The more interesting part is the time domain for the transfer funtion \$h(t)\$.
I will employ Laplace transforms
$$\mathcal{L}\left[Ae^{-at}\cos{{\omega}t}\right]=\frac{A(s+a)}{(s+a)^2+\omega^2}\tag{5}\label{5}$$
and
$$\mathcal{L}\left[Be^{-at}\sin{{\omega}t}\right]=\frac{B\omega}{(s+a)^2+\omega^2}\tag{6}\label{6}$$
or
$$\mathcal{L}\left[Ae^{-at}\cos{{\omega}t}+Be^{-at}\sin{{\omega}t}\right] = \frac{A(s+a) + B\omega}{(s+a)^2+\omega^2}\tag{7}\label{7}$$
Completing the square for \$D(s)\$ in \$H(s)=\frac{N(s)}{D(s)}\$ to reach the form of \$\ref{7}\$
$$H(s)=10\frac{(s+10) + 290}{(s+10)^2+49900}\tag{8}$$
And figuring out \$B\$ with \$A=1\$ gives
$$B=\frac{290}{\sqrt{49900}}\tag{9}$$
Then the time domain of the transfer by substituting values into the left side of \$\ref{7}\$ becomes
$$h(t)=10e^{-10t}\left(\cos(\sqrt{49900}t)+B\sin(\sqrt{49900}t)\right)\tag{10}$$
Letting \$C=\sqrt{A^2+B^2}=\sqrt{1^2+B^2}\$
$$h(t)=10e^{-10t}C\left(\frac{1}{C}\cos(\sqrt{49900}t)+\frac{B}{C}\sin(\sqrt{49900}t)\right)\tag{11}$$
Then \$\cos{\phi}=\frac{1}{C}\$ and \$\sin{\phi}=\frac{B}{C}\$
$$h(t)=10e^{-10t}C\left(\cos{\phi}\cos(\sqrt{49900}t)+\sin{\phi}\sin(\sqrt{49900}t)\right)\tag{12}$$
Using a trigonometric identity becomes
$$h(t)=16.3871e^{-10t}\cos(223.3831t - 0.9144)\tag{13}$$