Đáp án:
$$c)\,\,\left[ \matrix{
x = k2\pi \hfill \cr
x = {\pi \over 9} + {{k2\pi } \over 3} \hfill \cr} \right.\,\,\left( {k \in Z} \right)$$
Giải thích các bước giải:
$$\eqalign{
& c)\,\,{{\sin x - \sin 2x} \over {1 - \cos x - 2{{\sin }^2}x}} = \sqrt 3 \cr
& \Leftrightarrow \sin x - \sin 2x = \sqrt 3 \left( {\cos 2x - \cos x} \right) \cr
& \Leftrightarrow \sin x + \sqrt 3 \cos x = \sin 2x + \sqrt 3 \cos 2x \cr
& \Leftrightarrow {1 \over 2}\sin x + {{\sqrt 3 } \over 2}\cos x = {1 \over 2}\sin 2x + {{\sqrt 3 } \over 2}\cos 2x \cr
& \Leftrightarrow \sin \left( {x + {\pi \over 3}} \right) = \sin \left( {2x + {\pi \over 3}} \right) \cr
& \Leftrightarrow \left[ \matrix{
2x + {\pi \over 3} = x + {\pi \over 3} + k2\pi \hfill \cr
2x + {\pi \over 3} = \pi - x - {\pi \over 3} + k2\pi \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = k2\pi \hfill \cr
x = {\pi \over 9} + {{k2\pi } \over 3} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$
