1 câu trả lời
Đáp án:
GTNN$y=-\sqrt3$
$\Leftrightarrow x=-\dfrac{\pi}3+k2\pi$ $(k\in\mathbb Z)$
Giải thích các bước giải:
Ta có: $y=\sin x+\sin\left({x-\dfrac{\pi}3}\right)$
$\Leftrightarrow y=\sin x+\dfrac12\sin x-\dfrac{\sqrt3}2\cos x$
$\Leftrightarrow \dfrac y{\sqrt3}=\dfrac{\sqrt3}2\sin x-\dfrac12\cos x$
$\Leftrightarrow \dfrac y{\sqrt3}=\sin\left({x-\dfrac{\pi}6}\right)\ge-1$
$\Rightarrow y\ge-\sqrt3$, GTNN$y=-\sqrt3\Leftrightarrow \sin\left({x-\dfrac{\pi}6}\right)=-1$
$\Leftrightarrow x-\dfrac{\pi}6=-\dfrac{\pi}2+k2\pi\Leftrightarrow x=-\dfrac{\pi}3+k2\pi$ $(k\in\mathbb Z)$.