Solve $\left(1\right)$ for $\omega\space$:
$$\frac{\text{P}_\text{out}\left(t\right)}{\text{P}_\text{in}\left(t\right)}=\frac{1}{2}\tag1$$
When:
$$\text{U}_\text{in}\left(t\right)=\hat{\text{u}}\cdot\cos\left(\omega\cdot t+\varphi\right)\tag2$$
And all the variables are real numbers.
My work:
I know that:
$$\frac{\text{P}_\text{out}\left(t\right)}{\text{P}_\text{in}\left(t\right)}=\frac{\text{U}_\text{out}\left(t\right)\cdot\text{I}_\text{out}\left(t\right)}{\text{U}_\text{in}\left(t\right)\cdot\text{I}_\text{in}\left(t\right)}=\frac{\text{U}_\text{out}\left(t\right)\cdot\text{I}_\text{in}\left(t\right)}{\text{U}_\text{in}\left(t\right)\cdot\text{I}_\text{in}\left(t\right)}=\frac{\text{U}_\text{out}\left(t\right)}{\text{U}_\text{in}\left(t\right)}\tag3$$
I also know:
$$\text{U}_\text{out}\left(t\right)=\text{U}_{\text{R}_2}\left(t\right)+\text{U}_\text{C}\left(t\right)\space\Longleftrightarrow\space\text{U}_\text{out}'\left(t\right)=\text{I}_\text{in}'\left(t\right)\cdot\text{R}_2+\text{I}_\text{in}\left(t\right)\cdot\frac{1}{\text{C}}\tag4$$
And:
$$\text{U}_\text{in}\left(t\right)=\text{U}_{\text{R}_1}\left(t\right)+\text{U}_{\text{R}_2}\left(t\right)+\text{U}_\text{C}\left(t\right)\space\Longleftrightarrow\space$$ $$\text{U}_\text{in}'\left(t\right)=\text{I}_\text{in}'\left(t\right)\cdot\text{R}_1+\text{I}_\text{in}'\left(t\right)\cdot\text{R}_2+\text{I}_\text{in}\left(t\right)\cdot\frac{1}{\text{C}}\tag5$$
We can solve equation $\left(5\right)$:
$$-\hat{\text{u}}\cdot\omega\cdot\sin\left(\omega\cdot t+\varphi\right)=\text{I}_\text{in}'\left(t\right)\cdot\left(\text{R}_1+\text{R}_2\right)+\text{I}_\text{in}\left(t\right)\cdot\frac{1}{\text{C}}\space\Longleftrightarrow\space$$ $$\text{I}_\text{in}\left(t\right)=-\hat{\text{u}}\cdot\omega\cdot\frac{\exp\left(-\frac{t}{\text{C}\cdot\left(\text{R}_1+\text{R}_2\right)}\right)}{\text{R}_1+\text{R}_2}\int\exp\left(\frac{t}{\text{C}\cdot\left(\text{R}_1+\text{R}_2\right)}\right)\cdot\sin\left(\omega\cdot t+\varphi\right)\space\text{d}t\tag6$$
$\color{red}{\text{how to continue?}}$
Integral on the right is of the form:
$$I = \int e^{at} \sin (bt + c) dt $$
Do the following:
$$u = e^{at}, dv=\sin (bt + c), du = a e^{at}, v = - \frac{\cos (bt + c)}{b}$$
$$I = uv - \int v du $$
$$I = - \frac {e^{at} \cos (bt + c)}{b} + \frac{a}{b} \int e^{at} \cos (bt + c) dt$$
Now solve the integral shown above by applying the same procedure again:
$$u = e^{at}, dv=\cos (bt + c), du = a e^{at}, v = \frac{\sin (bt + c)}{b}$$
$$I = - \frac {e^{at} \cos (bt + c)} {b} + \frac {a}{b} (\frac{e^{at} \sin (bt + c)}{b} - \frac{a}{b} \int e^{at} \sin (bt + c) dt)$$
$$I = - \frac {e^{at} \cos (bt + c)} {b} + \frac {a}{b} (\frac{e^{at} \sin (bt + c)}{b} - \frac{a}{b} I)$$
Solve for $I$ and you get:
$$I = \int e^{at} \sin (bt + c) dt = \frac{e^{at}}{a^2+b^2} [a \sin (bt + c) - b \cos (bt + c)]$$