Giải phương trình: cos3x - sin3x = 1

cos3x – sin3x = 1

⇔ \(\frac{{\sqrt 2 }}{2}\cos 3x - \frac{{\sqrt 2 }}{2}\sin 3x = \frac{{\sqrt 2 }}{2}\)

⇔ \(\cos \left( {3x + \frac{\pi }{4}} \right) = \cos \left( {\frac{\pi }{4}} \right)\)

⇔ \(\left[ \begin{array}{l}3x + \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\3x + \frac{\pi }{4} = - \frac{\pi }{4} + k2\pi \end{array} \right.\)

\[ \Leftrightarrow \left[ \begin{array}{l}x = k\frac{{2\pi }}{3}\\x = - \frac{\pi }{3} + k\frac{{2\pi }}{3}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\].