Đáp án:
\(x = \dfrac{{k2\pi }}{3};\,\,x = \dfrac{\pi }{4} + k\pi ;\,\,x = - \dfrac{\pi }{2} + k2\pi \,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
\,\,\,\,\,\cos x - \cos 2x = \sin 3x\\
\Leftrightarrow 2\sin \dfrac{{3x}}{2}\sin \dfrac{x}{2} = 2\sin \dfrac{{3x}}{2}\cos \dfrac{{3x}}{2}\\
\Leftrightarrow 2\sin \dfrac{{3x}}{2}\left( {\sin \dfrac{x}{2} - \cos \dfrac{{3x}}{2}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \dfrac{{3x}}{2} = 0\\
\sin \dfrac{x}{2} = \cos \dfrac{{3x}}{2} = \sin \left( {\dfrac{\pi }{2} - \dfrac{{3x}}{2}} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{3x}}{2} = k\pi \\
\dfrac{x}{2} = \dfrac{\pi }{2} - \dfrac{{3x}}{2} + k2\pi \\
\dfrac{x}{2} = \pi - \dfrac{\pi }{2} + \dfrac{{3x}}{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{4} + k\pi \\
x = - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\,\,\left( {k \in Z} \right)
\end{array}\)
