Đáp án: $MinY=-\sqrt{3}, MaxY=\sqrt{3}$
Giải thích các bước giải:
Ta có:
$y=\cos(x)+\cos(x-\dfrac{\pi}{3})$
$\to y=\cos x+\cos x\cdot\cos(\dfrac{\pi}{3})+\sin x\cdot \sin(\dfrac{\pi}{3})$
$\to y=\cos x+\dfrac12\cos x+\dfrac{\sqrt{3}}{2}\sin(x)$
$\to y=\dfrac32\cos(x)+\dfrac{\sqrt{3}}{2}\sin x$
$\to y^2=(\dfrac32\cos(x)+\dfrac{\sqrt{3}}{2}\sin x)^2$
$\to y^2\le ((\dfrac32)^2+(\dfrac{\sqrt{3}}{2})^2)(\cos^2x+\sin^2x)$
4\to y^2\le 3$
$\to -\sqrt{3}\le y\le \sqrt{3}$
$\to MinY=-\sqrt{3}, MaxY=\sqrt{3}$
Ta có $Max Y$ xảy ra khi $\dfrac{\cos x}{\dfrac32}=\dfrac{\sin x}{\dfrac{\sqrt{3}}{2}}$
$\to \dfrac{\sin x}{\cos x}=\dfrac{1}{\sqrt{3}}$
$\to\tan x=\dfrac{1}{\sqrt{3}}$
$Max Y$ xảy ra khi $\dfrac{\cos x}{\dfrac32}=-\dfrac{\sin x}{\dfrac{\sqrt{3}}{2}}$
$\to \dfrac{\sin x}{\cos x}=-\dfrac{1}{\sqrt{3}}$
$\to\tan x=-\dfrac{1}{\sqrt{3}}$
