Đáp án:
`\max=\sqrt{2}` khi `x=\frac{\pi}4+k2\pi(k\in\mathbb Z)`
`\min=-\sqrt{2}` khi `x=-\frac{3\pi}4+k2\pi(k\in\mathbb Z)`
Lời giải:
`sinx + cosx`
`=\sqrt2(\sin x.\sqrt2/2 + \cos x .\sqrt2/2)`
`=\sqrt2(\sin x.\cos \frac{\pi}{4} + \cos x. sin \frac{\pi}{4} )`
`= \sqrt2sin(x + \pi/4) `
Ta có:
`-1 ≤ \sin (x + \pi/4) ≤1`
`⇔ -\sqrt2 ≤ \sqrt2sin(x + \pi/4) ≤\sqrt2`
`\max=\sqrt{2}` khi `\sin (x+\frac{\pi}{4})=1 \Leftrightarrow (x+\frac{\pi}{4})=\frac{\pi}{2}+k2\pi`
`\Leftrightarrow x=\frac{\pi}4+k2\pi(k\in\mathbb Z)`
`\min=-\sqrt{2}` khi `\sin (x+\frac{\pi}{4})=-1 \Leftrightarrow (x+\frac{\pi}{4})=-\frac{\pi}{2}+k2\pi` `\Leftrightarrow x=-\frac{3\pi}4+k2\pi(k\in\mathbb Z)`
