Giải phương trình \(\sqrt 3 \sin 2x + \cos 2x = 2\cos x\).
Giải chi tiết:
Ta có:
\(\begin{array}{l}\,\,\,\,\,\sqrt 3 \sin 2x + \cos 2x = 2\cos x \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin 2x + \dfrac{1}{2}\cos 2x = \cos x\\ \Leftrightarrow \sin \dfrac{\pi }{3}.\sin 2x + \cos \dfrac{\pi }{3}.\cos 2x = \cos x\\ \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{3}} \right) = \cos x \Leftrightarrow \left[ \begin{array}{l}2x - \dfrac{\pi }{3} = x + k2\pi \\2x - \dfrac{\pi }{3} = - x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{3} + k2\pi \\x = \dfrac{\pi }{9} + k\dfrac{{2\pi }}{3}\end{array} \right.\,\,\left( {k \in Z} \right)\end{array}\)
Vậy, phương trình đã cho có nghiệm \(x = \dfrac{\pi }{3} + k2\pi ,\,\,x = \dfrac{\pi }{9} + k\dfrac{{2\pi }}{3};\,\,k \in Z\).