Đáp án:

$b)x =  - \dfrac{{5\pi }}{6} + k2\pi ;x = \dfrac{{5\pi }}{{18}} + k\dfrac{{2\pi }}{3}\left( {k \in Z} \right)$

$c)x = \pi  + k2\pi ;x = \dfrac{{2\pi }}{9} + k\dfrac{{2\pi }}{3}\left( {k \in Z} \right)$

$d)x =  - \dfrac{\pi }{{18}} + k\dfrac{{2\pi }}{3}\left( {k \in Z} \right)$

Giải thích các bước giải:

$\begin{array}{l}
b)\sin x - \sqrt 3 \cos x = 2\cos 2x\\
 \Leftrightarrow \dfrac{1}{2}\sin x - \dfrac{{\sqrt 3 }}{2}\cos x = \cos 2x\\
 \Leftrightarrow \cos \left( {x - \dfrac{{5\pi }}{6}} \right) = \cos 2x\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = x - \dfrac{{5\pi }}{6} + k2\pi \\
2x =  - x + \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{{5\pi }}{6} + k2\pi \\
x = \dfrac{{5\pi }}{{18}} + k\dfrac{{2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$ 

$\begin{array}{l}
c)\sin x + \sin 2x - \sqrt 3 \left( {\cos x + \cos 2x} \right) = 0\\
 \Leftrightarrow \sin x - \sqrt 3 \cos x = \sqrt 3 \cos 2x - \sin 2x\\
 \Leftrightarrow \dfrac{1}{2}\sin x - \dfrac{{\sqrt 3 }}{2}\cos x = \dfrac{{\sqrt 3 }}{2}\cos 2x - \dfrac{1}{2}\sin 2x\\
 \Leftrightarrow \cos \left( {x - \dfrac{{5\pi }}{6}} \right) = \cos \left( {2x + \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{6} = x - \dfrac{{5\pi }}{6} + k2\pi \\
2x + \dfrac{\pi }{6} =  - x + \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \pi  + k2\pi \\
x = \dfrac{{2\pi }}{9} + k\dfrac{{2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$

$\begin{array}{l}
d)\dfrac{{\left( {1 - 2\sin x} \right)\cos x}}{{\left( {1 + 2\sin x} \right)\left( {1 - \sin x} \right)}} = \sqrt 3 \left( {DK:\sin x \ne \left\{ {\dfrac{{ - 1}}{2};1} \right\}} \right)\\
 \Leftrightarrow \dfrac{{\cos x - 2\sin x\cos x}}{{1 - 2{{\sin }^2}x + \sin x}} = \sqrt 3 \\
 \Leftrightarrow \dfrac{{\cos x - \sin 2x}}{{\cos 2x + \sin x}} = \sqrt 3 \\
 \Leftrightarrow \cos x - \sin 2x = \sqrt 3 \cos 2x + \sqrt 3 \sin x\\
 \Leftrightarrow \dfrac{1}{2}\cos x - \dfrac{{\sqrt 3 }}{2}\sin x = \dfrac{{\sqrt 3 }}{2}\cos 2x + \dfrac{1}{2}\sin 2x\\
 \Leftrightarrow \cos \left( {x + \dfrac{\pi }{3}} \right) = \cos \left( {2x - \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{3} = 2x - \dfrac{\pi }{6} + k2\pi \\
x + \dfrac{\pi }{3} =  - 2x + \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \left( l \right)\\
x =  - \dfrac{\pi }{{18}} + k\dfrac{{2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow x =  - \dfrac{\pi }{{18}} + k\dfrac{{2\pi }}{3}\left( {k \in Z} \right)
\end{array}$