Đáp án:
$\begin{cases}miny = \dfrac{1}{2} \Leftrightarrow x = -\dfrac{\pi}{4} + k\pi \\maxy = \dfrac{3}{2} \Leftrightarrow x = \dfrac{\pi}{4} + k\pi\end{cases}\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$y = 1 -\sin x\cos x$
$= 1 - \dfrac{1}{2}.(2\sin x \cos x)$
$=1 - \dfrac{1}{2}\sin2x$
Ta có: $-1 \leq \sin2x \leq 1$
$\Leftrightarrow -\dfrac{1}{2} \leq -\dfrac{1}{2}\sin2x \leq \dfrac{1}{2}$
$\Leftrightarrow \dfrac{1}{2} \leq 1-\dfrac{1}{2}\sin2x \leq \dfrac{3}{2}$
Hay $\dfrac{1}{2} \leq y \leq \dfrac{3}{2}$
Vậy $miny = \dfrac{1}{2} \Leftrightarrow \sin2x = 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\pi$
$maxy = \dfrac{3}{2} \Leftrightarrow \sin2x = -1 \Leftrightarrow x = -\dfrac{\pi}{4} + k\pi$ $(k \in \Bbb Z)$
