Đáp án:
$\begin{cases}\min B = -\sqrt2 \Leftrightarrow x = -\dfrac{3\pi}{4} + k2\pi\\\max B =\sqrt2 \Leftrightarrow x =\dfrac{\pi}{4}+ k2\pi\end{cases}\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$B = \sin x + \cos x$
$\to B = \sqrt2\sin\left(x + \dfrac{\pi}{4}\right)$
Ta có:
$-1 \leq \sin\left(x + \dfrac{\pi}{4}\right) \leq 1$
$\to - \sqrt2 \leq \sqrt2\sin\left(x + \dfrac{\pi}{4}\right) \leq \sqrt2$
$\to -\sqrt2 \leq B \leq \sqrt2$
Vậy $\min B = -\sqrt2 \Leftrightarrow x = -\dfrac{3\pi}{4} + k2\pi$
$\max B =\sqrt2 \Leftrightarrow x =\dfrac{\pi}{4}+ k2\pi\quad (k\in\Bbb Z)$
