sin³x + cos³x = 1 – \(\frac{1}{2}\sin 2x\)

⇔ (sinx + cosx)(sin²x – sinxcosx + cos²x) = 1 – \(\frac{1}{2}\sin 2x\)

⇔ (sinx + cosx)(1 – sinxcosx) = 1 – \(\frac{1}{2}\sin 2x\)

⇔ (sinx + cosx)\(\left( {1 - \frac{1}{2}\sin 2x} \right) = 1 - \frac{1}{2}\sin 2x\)

⇔ \(\left( {1 - \frac{1}{2}\sin 2x} \right)\left( {\sin x + \cos x - 1} \right) = 0\)

⇔ \(\left[ \begin{array}{l}\sin 2x = 2 > 1\left( L \right)\\\sin x + \cos x - 1 = 0\end{array} \right.\)

⇔ sinx + cosx = 1

⇔ \(\cos \left( {x - \frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2} = \cos \left( {\frac{\pi }{4}} \right)\)

⇔ \(x - \frac{\pi }{4} = \pm \frac{\pi }{4} + k2\pi \)

⇔ \(\left[ \begin{array}{l}x = k2\pi \\x = \frac{\pi }{2} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\).