sin3x + cos3x = sin2x + 1 + sinx + cosx
⇔ (sinx + cosx)(sin2x + cos2x - sinx.cosx) = 2sinxcosx + sin2x + cos2x + sinx + cosx
⇔ (sinx + cosx)(1 - sinx . cosx) = (sinx + cosx)2 + (sinx + cosx)
⇔ (sinx + cosx)(1 - sinx.cosx - sinx - cosx - 1) = 0
⇔ (sinx + cosx)(sinx + cosx + sinx.cosx) = 0
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\sinx+cosx+sinx.cosx=0\left(2\right)\end{matrix}\right.\)
(1) ⇔ \(\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\)
⇔ \(x+\dfrac{\pi}{4}=k\pi\)
⇔ \(x=-\dfrac{\pi}{4}+k\pi\)
(2) ⇔ \(\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+\dfrac{1}{2}sin2x=0\)
⇔ \(\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)-\dfrac{1}{2}cos\left(2x+\dfrac{\pi}{2}\right)=0\)
⇔ \(\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)-\dfrac{1}{2}.\left[1-2sin^2\left(x+\dfrac{\pi}{4}\right)\right]=0\)
⇔ \(\dfrac{1}{4}sin^2\left(x+\dfrac{\pi}{4}\right)+\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)-\dfrac{1}{2}=0\)
⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{10}-2\sqrt{2}\\sin\left(x+\dfrac{\pi}{4}\right)=-\sqrt{10}-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{10}-2\sqrt{2}\)
