Đáp án:
$\min y = -4;\,\max y = -\dfrac74$
Giải thích các bước giải:
$y =\cos^2x -\sin x -3$
$\to y = 1 -\sin^2x -\sin x -3$
$\to y = -\left(\sin^2x +2\cdot\dfrac12\sin x +\dfrac14\right) -\dfrac74$
$\to y = -\left(\sin x +\dfrac12\right)^2 -\dfrac74$
Ta có:
$\quad -1\leq \sin x \leq 1$
$\to -\dfrac12 \leq \sin x +\dfrac12\leq \dfrac32$
$\to 0 \leq \left(\sin x +\dfrac12\right)^2\leq \dfrac94$
$\to -\dfrac94 \leq -\left(\sin x +\dfrac12\right)^2 \leq 0$
$\to -4 \leq -\left(\sin x +\dfrac12\right)^2 -\dfrac74 \leq -\dfrac74$
$\to -4 \leq y \leq -\dfrac74$
Vậy $\min y = -4;\,\max y = -\dfrac74$
