Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{\pi }{{28}} + \dfrac{{k2\pi }}{7}\\
x = \dfrac{{5\pi }}{{12}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\sin 5x - \cos 5x = - 2\sqrt 2 \cos x.\sin x\\
\to \sqrt 2 \sin \left( {5x - \dfrac{\pi }{4}} \right) = - \sqrt 2 \sin 2x\\
\to \sin \left( {5x - \dfrac{\pi }{4}} \right) = - \sin 2x\\
\to \sin \left( {5x - \dfrac{\pi }{4}} \right) + \sin 2x = 0\\
\to 2\sin \left( {\dfrac{{5x - \dfrac{\pi }{4} + 2x}}{2}} \right).\cos \left( {\dfrac{{5x - \dfrac{\pi }{4} - 2x}}{2}} \right) = 0\\
\to 2\sin \left( {\dfrac{7}{2}x - \dfrac{\pi }{8}} \right).\cos \left( {\dfrac{3}{2}x - \dfrac{\pi }{8}} \right) = 0\\
\to \left[ \begin{array}{l}
\sin \left( {\dfrac{7}{2}x - \dfrac{\pi }{8}} \right) = 0\\
\cos \left( {\dfrac{3}{2}x - \dfrac{\pi }{8}} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{7}{2}x - \dfrac{\pi }{8} = k\pi \\
\dfrac{3}{2}x - \dfrac{\pi }{8} = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{{28}} + \dfrac{{k2\pi }}{7}\\
x = \dfrac{{5\pi }}{{12}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
