Đáp án:
\[\left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
- 2\sin x.\sin y = \cos \left( {x + y} \right) - \cos \left( {x - y} \right)\\
\cos 2x\left( {1 + 2\cos 2x} \right) = 1 + \sin x\left( {1 - 2\sin 3x} \right) + \sqrt 3 \cos x\\
\Leftrightarrow \cos 2x + 2{\cos ^2}2x = 1 + \sin x - 2\sin x.\sin 3x + \sqrt 3 \cos x\\
\Leftrightarrow \cos 2x + \left( {2{{\cos }^2}2x - 1} \right) = \sin x + \cos \left( {x + 3x} \right) - \cos \left( {x - 3x} \right) + \sqrt 3 \cos x\\
\Leftrightarrow \cos 2x + \cos 4x = \sin x + \cos 4x - \cos \left( { - 2x} \right) + \sqrt 3 \cos x\\
\Leftrightarrow \sin x + \sqrt 3 \cos x = 2\cos 2x\\
\Leftrightarrow \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x = \cos 2x\\
\Leftrightarrow \cos x.\cos \dfrac{\pi }{6} + \sin x.\sin \dfrac{\pi }{6} = \cos 2x\\
\Leftrightarrow \cos \left( {x - \dfrac{\pi }{6}} \right) = \cos 2x\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{6} = 2x + k2\pi \\
x - \dfrac{\pi }{6} = - 2x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Vậy nghiệm của phương trình đã cho là \(\left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\)
