Đáp án:
$\min y = -3;\, \max y = \dfrac18$
Giải thích các bước giải:
$y =\cos2x + \sin x - 1$
$\to y = 1 - 2\sin^2x +\sin x -1$
$\to y = -2\sin^2x +\sin x$
$\to y = -2\left(\sin^2x -2.\dfrac{1}{4}\sin x +\dfrac{1}{16}\right) +\dfrac{1}{8}$
$\to y = -2\left(\sin x -\dfrac{1}{4}\right)^2 +\dfrac{1}{8}$
Ta có:
$-1 \leq \sin x \leq 1$
$\to -\dfrac{5}{4} \leq \sin x -\dfrac14 \leq \dfrac{3}{4}$
$\to 0 \leq \left(\sin x -\dfrac{1}{4}\right)^2 \leq \dfrac{25}{16}$
$\to -\dfrac{25}{8} \leq -2\left(\sin x -\dfrac{1}{4}\right)^2 \leq 0$
$\to -3 \leq -2\left(\sin x -\dfrac{1}{4}\right)^2 \leq \dfrac18$
$\to -3 \leq y \leq\dfrac18$
Vậy $\min y = -3;\, \max y = \dfrac18$
