Giải thích các bước giải:
1.Ta có :
$y=2-\sin x-\cos x=2-\sqrt{2}.\sin (x+\dfrac{\pi}{4})$
$\to 2-\sqrt{2}\le y\le 2+\sqrt{2}$
$\to Max y=2+\sqrt{2}\to \sin (x+\dfrac{\pi}{4})=-1$
$Min y=2-\sqrt{2}\to\sin (x+\dfrac{\pi}{4})=1$
2.$\cos 2x=\dfrac{-1}{2}$
$\to 2x=\dfrac{2\pi}{3}+k2\pi$
$\to x=\dfrac{\pi}{3}+k\pi$
Hoặc $2x=\dfrac{4\pi}{3}+k2\pi\to x=\dfrac{2\pi}{3}+k\pi$
Vì $0\le x\le 5\pi$
$\rightarrow 0\le\dfrac{\pi}{3}+k\pi \le 5\pi\rightarrow 0\le k\le 4$
Hoặc $ 0\le\dfrac{2\pi}{3}+k\pi \le 5\pi\to 0\le k\le 4$
$\Rightarrow$Có 10 nghiệm
