tìm GTLN, GTNN của hàm số y = cos(2x-pi/6) + cos2x -3. Giúp mình với ạ 31/07/2021 Bởi Genesis tìm GTLN, GTNN của hàm số y = cos(2x-pi/6) + cos2x -3. Giúp mình với ạ
$y = \cos\left(2x – \dfrac{\pi}{6}\right) + \cos2x – 3$ $= \cos2x.\cos\dfrac{\pi}{6} + \sin2x.\sin\dfrac{\pi}{6} + \cos2x -3$ $= \left(1 + \dfrac{\sqrt3}{2}\right)\cos2x + \dfrac{1}{2}\sin2x – 3$ $\Leftrightarrow y + 3 = \left(1 + \dfrac{\sqrt3}{2}\right)\cos2x + \dfrac{1}{2}\sin2x$ $\Rightarrow (y+3)^2 \leq \left(1 + \dfrac{\sqrt3}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2 = 2 + \sqrt3$ $\Rightarrow -\sqrt{2 + \sqrt3} \leq y + 3\leq \sqrt{2 + \sqrt3}$ $\Leftrightarrow -\sqrt{2 + \sqrt3} – 3 \leq y \leq \sqrt{2 + \sqrt3} – 3$ Vậy $\min y = -\sqrt{2 + \sqrt3} – 3$ $\max y = \sqrt{2 + \sqrt3} – 3$ Bình luận
$y = \cos\left(2x – \dfrac{\pi}{6}\right) + \cos2x – 3$
$= \cos2x.\cos\dfrac{\pi}{6} + \sin2x.\sin\dfrac{\pi}{6} + \cos2x -3$
$= \left(1 + \dfrac{\sqrt3}{2}\right)\cos2x + \dfrac{1}{2}\sin2x – 3$
$\Leftrightarrow y + 3 = \left(1 + \dfrac{\sqrt3}{2}\right)\cos2x + \dfrac{1}{2}\sin2x$
$\Rightarrow (y+3)^2 \leq \left(1 + \dfrac{\sqrt3}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2 = 2 + \sqrt3$
$\Rightarrow -\sqrt{2 + \sqrt3} \leq y + 3\leq \sqrt{2 + \sqrt3}$
$\Leftrightarrow -\sqrt{2 + \sqrt3} – 3 \leq y \leq \sqrt{2 + \sqrt3} – 3$
Vậy $\min y = -\sqrt{2 + \sqrt3} – 3$
$\max y = \sqrt{2 + \sqrt3} – 3$