y = 3cosx + 4sinx tim gtnn va gtln

Đáp án:

$\min y = – 5 \Leftrightarrow x = {\pi \over 2} – \alpha + k2\pi $

$\max y = 5 \Leftrightarrow  x = – {\pi \over 2} – \alpha + k2\pi $

$(k\in\mathbb Z)$

Lời giải:

$\eqalign{ & y = 3\cos x + 4\sin x \cr & y = 5\left( {{3 \over 5}\cos x + {4 \over 5}\sin x} \right) \cr & \text{Đặt }\left\{ \matrix{ \sin \alpha = {3 \over 5} \hfill \cr \cos \alpha = {4 \over 5} \hfill \cr} \right. \Rightarrow y = 5\left( {\sin \alpha \cos x + \cos \alpha \sin x} \right) = 5\sin \left( {\alpha + x} \right) \cr & \text{ta có: } – 1 \le \sin \left( {\alpha + x} \right) \le 1 \Leftrightarrow – 5 \le 5\sin \left( {\alpha + x} \right) \le 5 \cr & \Rightarrow – 5 \le y \le 5 \cr & \text{Vậy } \cr & \min y = – 5 \Leftrightarrow \sin \left( {\alpha + x} \right) = 1 \Leftrightarrow x + \alpha = {\pi \over 2} + k2\pi \Leftrightarrow x = {\pi \over 2} – \alpha + k2\pi \cr & \max y = 5 \Leftrightarrow \sin \left( {\alpha + x} \right) = – 1 \Leftrightarrow x + \alpha = – {\pi \over 2} + k2\pi \Leftrightarrow x = – {\pi \over 2} – \alpha + k2\pi \cr} $