Toán Giải phương trình : cosx – cox2x= sin3x 06/09/2021 By Aaliyah Giải phương trình : cosx – cox2x= sin3x
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: \(\begin{array}{l}\,\,\,\,\,\cos x – \cos 2x = \sin 3x\\ \Leftrightarrow 2\sin \dfrac{{3x}}{2}\sin \dfrac{x}{2} = 2\sin \dfrac{{3x}}{2}\cos \dfrac{{3x}}{2}\\ \Leftrightarrow 2\sin \dfrac{{3x}}{2}\left( {\sin \dfrac{x}{2} – \cos \dfrac{{3x}}{2}} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin \dfrac{{3x}}{2} = 0\\\sin \dfrac{x}{2} = \cos \dfrac{{3x}}{2} = \sin \left( {\dfrac{\pi }{2} – \dfrac{{3x}}{2}} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\dfrac{{3x}}{2} = k\pi \\\dfrac{x}{2} = \dfrac{\pi }{2} – \dfrac{{3x}}{2} + k2\pi \\\dfrac{x}{2} = \pi – \dfrac{\pi }{2} + \dfrac{{3x}}{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{{k2\pi }}{3}\\x = \dfrac{\pi }{4} + k\pi \\x = – \dfrac{\pi }{2} + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right)\end{array}\) Trả lời
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:
\(\begin{array}{l}
\,\,\,\,\,\cos x – \cos 2x = \sin 3x\\
\Leftrightarrow 2\sin \dfrac{{3x}}{2}\sin \dfrac{x}{2} = 2\sin \dfrac{{3x}}{2}\cos \dfrac{{3x}}{2}\\
\Leftrightarrow 2\sin \dfrac{{3x}}{2}\left( {\sin \dfrac{x}{2} – \cos \dfrac{{3x}}{2}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \dfrac{{3x}}{2} = 0\\
\sin \dfrac{x}{2} = \cos \dfrac{{3x}}{2} = \sin \left( {\dfrac{\pi }{2} – \dfrac{{3x}}{2}} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{3x}}{2} = k\pi \\
\dfrac{x}{2} = \dfrac{\pi }{2} – \dfrac{{3x}}{2} + k2\pi \\
\dfrac{x}{2} = \pi – \dfrac{\pi }{2} + \dfrac{{3x}}{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{4} + k\pi \\
x = – \dfrac{\pi }{2} + k2\pi
\end{array} \right.\,\,\left( {k \in Z} \right)
\end{array}\)