There was a problem regarding an outside temperature varying as a sine wave, after solving the ODE I got a function, which I later on differentiated to get the maximum and minimum temperature, however I need to figure out $t_{max}$ and $t_{min}$, this is what I got after differentitating:
$$\tan\left(\dfrac{\pi t}{12} \right)=\dfrac{\pi}{6}$$
However I am not sure how can I find the maximum and minimum $t$ here?
The original equation is:
$$T(t)=65-\dfrac{540}{36+\pi^{2}}\cos\left(\dfrac{\pi t}{12}\right)-\dfrac{90\pi}{36+\pi^{2}}\sin\left(\dfrac{\pi t}{12} \right) $$
The equation $$\tan\left(\dfrac{\pi t}{12} \right)=\dfrac{\pi}{6}$$ has many solutions.
Let $$ \alpha = \dfrac{\pi t}{12}$$
One solution to $tan (\alpha )= \pi/6$ is $\alpha = \tan ^{-1} (\pi /6)$ where $\alpha$ is in the first quadrant, which makes both $\sin (\alpha)$ and $\cos (\alpha )$ to be positive.
For this value of $\alpha $ we find $$t=\frac {12\alpha}{\pi }$$ is a minimizer of $T(t)$ because we are subtracting positive values from $65$
Thus $$t_{min} = \frac {12\alpha }{\pi}.$$
To find the maximizer we choose $\beta = \pi + \alpha$, where both $\sin (\beta )$ and $\cos (\beta )$ are negative.
Thus $$t_{max} = \frac {12\beta }{\pi}.$$