Đáp án:

$$\eqalign{ & b)\,\,\left[ \matrix{ x = k2\pi \hfill \cr x = {\pi \over 9} + {{k2\pi } \over 3} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr & c)\,\,\left[ \matrix{ x = - {\pi \over 6} + k2\pi \hfill \cr x = {{7\pi } \over {50}} + {{k2\pi } \over 5} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$

Giải thích các bước giải:

$$\eqalign{ & b)\,\,{{\sin x - \sin 2x} \over {1 - \cos x - 2{{\sin }^2}x}} = \sqrt 3 \cr & \Leftrightarrow \sin x - \sin 2x = \sqrt 3 \left( {\cos 2x - \cos x} \right) \cr & \Leftrightarrow \sin x + \sqrt 3 \cos x = \sin 2x + \sqrt 3 \cos 2x \cr & \Leftrightarrow {1 \over 2}\sin x + {{\sqrt 3 } \over 2}\cos x = {1 \over 2}\sin 2x + {{\sqrt 3 } \over 2}\cos 2x \cr & \Leftrightarrow \sin \left( {x + {\pi \over 3}} \right) = \sin \left( {2x + {\pi \over 3}} \right) \cr & \Leftrightarrow \left[ \matrix{ 2x + {\pi \over 3} = x + {\pi \over 3} + k2\pi \hfill \cr 2x + {\pi \over 3} = \pi - x - {\pi \over 3} + k2\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = k2\pi \hfill \cr x = {\pi \over 9} + {{k2\pi } \over 3} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr & c)\,\,\sqrt 3 \sin 2x - \cos 2x = 2\sin 3x \cr & \Leftrightarrow {{\sqrt 3 } \over 2}\sin 2x - {1 \over 2}\cos 2x = \sin 3x \cr & \Leftrightarrow \sin 2x\cos {\pi \over 6} - \cos 2x\sin {\pi \over 6} = \sin 3x \cr & \Leftrightarrow \sin \left( {2x - {\pi \over 6}} \right) = \sin 3x \cr & \Leftrightarrow \left[ \matrix{ 3x = 2x - {\pi \over 6} + k2\pi \hfill \cr 3x = \pi - 2x + {\pi \over 6} + k2\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = - {\pi \over 6} + k2\pi \hfill \cr x = {{7\pi } \over {50}} + {{k2\pi } \over 5} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$