Đáp án:
Giải thích các bước giải:
`1+sin 2x=sin x-cos x`
`⇔ 1+2sin xcos x=sin x-cos x\ (1)`
Đặt `t=sin x-cos x` (ĐK: `-\sqrt{2} \le x \le \sqrt{2})`
`⇒ sin x.cos x=\frac{1-t^2}{2}`
Thay vào `(1)`
`1+2.\frac{1-t^2}{2}=t`
`⇔ 1+1-t^2=t`
`⇔ t^2+t-2=0`
`⇔` \(\left[ \begin{array}{l}t=1\ (TM)\\t=-2\ (L)\end{array} \right.\)
`⇒ sin x-cos x=1`
`⇔ \sqrt{2}.sin(x-\frac{\pi}{4})=1`
`⇔ sin (x-\frac{\pi}{4})=\frac{1}{\sqrt{2}}`
`⇔` \(\left[ \begin{array}{l}x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})\\x-\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})\\x=\pi+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy ........
