Question

(1+ √2)(sinx+cosx)- sin2x= 1+√2 

Answer 1

\(\left(1+\sqrt{2}\right)\left(sinx+cosx\right)-sin2x=1+\sqrt{2}\)

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

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

⇔ \(\left(2+\sqrt{2}\right).sin\left(x+\dfrac{\pi}{4}\right)-2sin^2\left(x+\dfrac{\pi}{4}\right)-\sqrt{2}=0\)

⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=1\\sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)