Đáp án:
GTLN $y=\sqrt2$ khi $x=\dfrac{\pi}8+k\pi$
Lời giải:
$y=\cos 2x+\sin 2x=\sqrt2\sin x(2x+\dfrac{\pi}{4})$
Do $-1\le\sin x\le1$ $\forall x$
$\Rightarrow-1\le\sin x(2x+\dfrac{\pi}{4})\le1$
$\Rightarrow-\sqrt2\le\sqrt2\sin x(2x+\dfrac{\pi}{4})\le\sqrt2$
$\Rightarrow-\sqrt2\le y\le \sqrt2$
GTLN $y=\sqrt2$ khi $\sin (2x+\dfrac{\pi}{4})=1\Leftrightarrow 2x+\dfrac{\pi}4=\dfrac{\pi}2+k2\pi$
$\Leftrightarrow x=\dfrac{\pi}8+k\pi$.
