Đáp án:
\(x = \dfrac{\pi }{6} + k2\pi ,x = \dfrac{{5\pi }}{6} + k2\pi ,x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3},k \in \mathbb{Z}\)
Giải thích các bước giải:
\(\begin{array}{l}\sqrt 3 \cos 2x\left( {2\sin x - 1} \right) + 2\cos x\left( {2{{\sin }^2}x + 1} \right) = 3\sin 2x\\ \Leftrightarrow \sqrt 3 \cos 2x\left( {2\sin x - 1} \right) + 2\cos x\left( {2{{\sin }^2}x + 1} \right) - 6\sin x\cos x = 0\\ \Leftrightarrow \sqrt 3 \cos 2x\left( {2\sin x - 1} \right) + 2\cos x\left( {2{{\sin }^2}x + 1 - 3\sin x} \right) = 0\\ \Leftrightarrow \sqrt 3 \cos 2x\left( {2\sin x - 1} \right) + 2\cos x\left( {\sin x - 1} \right)\left( {2\sin x - 1} \right) = 0\\ \Leftrightarrow \left( {2\sin x - 1} \right)\left( {\sqrt 3 \cos 2x + 2\sin x\cos x - 2\cos x} \right) = 0\\ \Leftrightarrow \left( {2\sin x - 1} \right)\left( {\sqrt 3 \cos 2x + \sin 2x - 2\cos x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}2\sin x - 1 = 0\,\,\left( 1 \right)\\\sqrt 3 \cos 2x + \sin 2x - 2\cos x = 0\,\left( 2 \right)\end{array} \right.\end{array}\)
\(\left( 1 \right) \Leftrightarrow \sin x = \dfrac{1}{2} \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \end{array} \right.,k \in \mathbb{Z}\)
\(\begin{array}{l}\left( 2 \right) \Leftrightarrow \sqrt 3 \cos 2x + \sin 2x = 2\cos x\\ \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos 2x + \dfrac{1}{2}\sin 2x = \cos x\\ \Leftrightarrow \cos \dfrac{\pi }{6}\cos 2x + \sin \dfrac{\pi }{6}\sin 2x = \cos x\\ \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{6}} \right) = \cos x\\ \Leftrightarrow \left[ \begin{array}{l}2x - \dfrac{\pi }{6} = x + k2\pi \\2x - \dfrac{\pi }{6} = - x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\end{array} \right.,k \in \mathbb{Z}\end{array}\)
Vậy pt có nghiệm \(x = \dfrac{\pi }{6} + k2\pi ,x = \dfrac{{5\pi }}{6} + k2\pi ,x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3},k \in \mathbb{Z}\)
