The problem states that the outside temperature varies as a sine wave with a minimum of $16^\circ C$ at $2$ a.m. and a mximum of $32^\circ C$ at $2$ p.m. . These are the steps I took for solving:
The equation for the outside temperature is:
$$M(t) = M_{0} +B \cos \left( wt \right)$$
To get $M_{0}$:
$$M_{0}=\dfrac{32+16}{2}=24$$
To find value of $B$, we set $M(0)=16$:
$$16=24+B$$ $$B=-8$$
Therefore,
$$M(t) = 24 -8 \cos \left(\dfrac{\pi t}{12} \right)$$
Now I plug $M(t)$ into the newton's law of cooling: $$\dfrac{d T}{dt}=K\left[M(t)-T(t) \right]$$ The problem mentioned that the time constant is $1$, therefore $K=1$, $$\dfrac{d T}{dt}=K\left[M(t)-T(t) \right]$$ $$\dfrac{d T}{dt}=24 -8 \cos \left(\dfrac{\pi t}{12} \right)-T(t)$$ $$\dfrac{d T}{dt}+T(t)=24 -8 \cos \left(\dfrac{\pi t}{12} \right)$$
Solving this ODE using integrating factor, we will have,
$$e^{t} T(t) = \int 24e^{t}-e^{t}\cos \left(\dfrac{\pi t}{12} \right) \; dt$$ $$e^{t} T(t)=-\dfrac{12\mathrm{e}^t\left({\pi}\left(\sin\left(\frac{{\pi}t}{12}\right)-2{\pi}\right)+12\cos\left(\frac{{\pi}t}{12}\right)-288\right)}{{\pi}^2+144}+C_{1}$$
We will neglect $e^{t}$ as said in the problem, we will have:
$$T(t)=24-\dfrac{12\pi}{\pi^{2}+144}\sin \dfrac{\pi t}{12} -\dfrac{144}{\pi^{2} + 144}\cos \dfrac{\pi t}{12}$$
To get lowest and highest temperature, we will find the derivative where $T'(t)=0$:
$$T'(t)=\dfrac{144}{\pi^{2} + 144} \sin \dfrac{\pi t}{12} - \dfrac{12 \pi}{\pi^{2}+144} \sin \dfrac{\pi t}{12}$$
$$\dfrac{144}{\pi^{2} + 144} \sin \dfrac{\pi t}{12}=\dfrac{12 \pi}{\pi^{2}+144} \sin \dfrac{\pi t}{12}$$
$$\tan \dfrac{\pi t}{12} = \dfrac{\pi}{12}$$
Now this is pretty much where I am stuck. I don't know how did the instructor get the maximum and minimum $t$ from this. We need the $t_{min}$ and $t_{max}$ to get the highest and lowest temperature.
If I properly understood, what you want is to solve for $t$ $$\tan \left(\frac{\pi t}{12}\right)=\frac{\pi }{12}$$ Just use the inverse function to get $$t_1=\frac{12 }{\pi }\tan ^{-1}\left(\frac{\pi }{12}\right)\qquad \text{and}\qquad t_2=\frac{12 }{\pi }\left(\tan ^{-1}\left(\frac{\pi }{12}\right)+\pi \right)$$
Now, may be, the question is to evaluate $\tan ^{-1}\left(\frac{\pi }{12}\right)$. SInce the angle is small, you could use the Taylor expansion $$\tan ^{-1}(x)=x-\frac{x^3}{3}+\frac{x^5}{5}+O\left(x^6\right)$$ Using it, you would find $$\tan ^{-1}\left(\frac{\pi }{12}\right)\approx 0.256064$$ (the exact value being $0.256053$).
Compute now $t_1$ and $t_2$.