Từ giả thiết, ta có:

\(\begin{array}{l}3\sin \left[ {\frac{\pi }{{182}}(t - 80)} \right] + 12 = 12 \Leftrightarrow \sin \left[ {\frac{\pi }{{182}}(t - 80)} \right] = 0\\ \Leftrightarrow \frac{\pi }{{182}}(t - 80) = k\pi  \Leftrightarrow t = 182k + 80,k \in \mathbb{Z}.\end{array}\)

Với \(t < 0 \le 365,\) ta có:

\(\begin{array}{l}0 < 182k + 80 \le 365 \Rightarrow  - \frac{{80}}{{182}} < k < \frac{{285}}{{182}}\\ \Rightarrow \left[ \begin{array}{l}k = 0\\k = 1\end{array} \right. \Rightarrow \left[ \begin{array}{l}t = 80\\t = 262\end{array} \right..\end{array}\)