Circuitous coffee container

Let $k$ be the constant of proportion in the Newton's law, i.e. $\frac{dT_{\text{cup}}}{dt}=-k(T_{\text{cup}}-T_{\text{env}})$ We need to find the value of $k$ . For this, we write the above equation as $\frac{dT_{\text{cup}}}{dt} + kT_{\text{cup}}=k(75+15 \cos \omega t)$ This is a linear differential equation. Let's try solution of the form $T_{\text{cup}}(t)=A\cos \omega t + B \sin \omega t + C$ For some constants $A,B,C$ . Substituting and equating the coefficients of both sides, we have $A=\frac{15k^2}{k^2+\omega^2}, B=\frac{15k\omega}{k^2+\omega^2}$ Hence, $\Delta T_{\text{cup}}=2\sqrt{A^2+B^2}$ Which yields, $k=\frac{\omega}{2\sqrt{2}}$ Finally, since the room temperature is constant, we have $\frac{dT_{\text{cup}}}{dt}=-k(T_{\text{cup}}-75)$ Solving this equation, with the supplied parameters, we get the result.