Ta có:

(1)\(cos\left(x\right)-sin\left(x\right)=\sqrt{2}.\dfrac{\sqrt{2}}{2}\left(cos\left(x\right)-sin\left(x\right)\right)\\ =\sqrt{2}.\left(\dfrac{\sqrt{2}}{2}.cos\left(x\right)-\dfrac{\sqrt{2}}{2}.sin\left(x\right)\right)\\ =\sqrt{2}.\left(sin\dfrac{\pi}{4}cos\left(x\right)-cos\dfrac{\pi}{4}.sin\left(x\right)\right)\\ =\sqrt{2}.sin\left(\dfrac{\pi}{4}-x\right)\)

(2) \(cos\left(x\right)+sin\left(x\right)=\sqrt{2}.\dfrac{\sqrt{2}}{2}\left(cos\left(x\right)+sin\left(x\right)\right)\\ =\sqrt{2}.\left(\dfrac{\sqrt{2}}{2}.cos\left(x\right)+\dfrac{\sqrt{2}}{2}.sin\left(x\right)\right)\\ =\sqrt{2}.\left(cos\dfrac{\pi}{4}cos\left(x\right)+sin\dfrac{\pi}{4}.sin\left(x\right)\right)\\ =\sqrt{2}.cos\left(x-\dfrac{\pi}{4}\right)\)

ADCT trên ta được:

\(sin\left(x\right)+\sqrt{2}.sin\left(\dfrac{\pi}{2}-x\right)=\sqrt{2}\\ \Leftrightarrow sin\left(x\right)+\sqrt{2}.sin\left(\dfrac{\pi}{4}-\left(x-\dfrac{\pi}{4}\right)\right)=\sqrt{2}\\ \Leftrightarrow sin\left(x\right)+cos\left(x-\dfrac{\pi}{4}\right)-sin\left(x-\dfrac{\pi}{4}\right)=\sqrt{2}\\ \sqrt{2}sin\left(x\right)+\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)+\sqrt{2}sin\left(\dfrac{\pi}{4}-x\right)=2\\ \Leftrightarrow\sqrt{2}sin\left(x\right)+cos\left(x\right)+sin\left(x\right)+cos\left(x\right)-sin\left(x\right)=2\\ \Leftrightarrow\sqrt{2}sin\left(x\right)+2cos\left(x\right)=2\)

Đến đây lại dùng cách trong sgk giải pt: a.sin(x) + b.cos(x) = c tìm ra x để thay nhá