Tìm GTLN,GTNN Y= √3 sinx-cosx

Đáp án:

$\begin{cases}miny = -2 \Leftrightarrow x = -\dfrac{\pi}{3} + k2\pi\\maxy = 2 \Leftrightarrow x = \dfrac{2\pi}{3} + k2\pi\end{cases} \quad (k \in \Bbb Z)$

Giải thích các bước giải:

$y = \sqrt3\sin x - \cos x$

$= 2\left(\dfrac{\sqrt3}{2}\sin x - \dfrac{1}{2}\cos x\right)$

$= 2\sin\left(x - \dfrac{\pi}{6}\right)$

Ta có:

$-1 \leq \sin\left(x - \dfrac{\pi}{6}\right) \leq 1$

$\Leftrightarrow -2 \leq 2\sin\left(x - \dfrac{\pi}{6}\right) \leq 2$

Hay $-2 \leq y \leq 2$

Vậy $miny = -2 \Leftrightarrow \sin\left(x - \dfrac{\pi}{6}\right) = -1 \Leftrightarrow x = -\dfrac{\pi}{3} + k2\pi$

$maxy = 2 \Leftrightarrow \sin\left(x - \dfrac{\pi}{6}\right) = 1 \Leftrightarrow x = \dfrac{2\pi}{3} + k2\pi \quad (k \in \Bbb Z)$