Phương trình \( {{\cos x - 2\sin x\cos x} \over {2...

D

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

ĐK:

\(  \eqalign{& \,\,\,\,\,\,2{\cos ^2}x + \sin x - 1 \ne 0 \cr & \Leftrightarrow 2 - 2{\sin ^2}x + \sin x - 1 \ne 0 \cr & \Leftrightarrow 2{\sin ^2}x - \sin x - 1 \ne 0 \cr & \Leftrightarrow \left( {2\sin x + 1} \right)\left( {\sin x - 1} \right) \ne 0 \cr & \Leftrightarrow \left\{ \matrix{\sin x \ne - {1 \over 2} \hfill \cr \sin x \ne 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{x \ne - {\pi \over 6} + k2\pi \hfill \cr x \ne {{7\pi } \over 6} + k2\pi \hfill \cr x \ne {\pi \over 2} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z } \right) \cr} \)

\(\eqalign{& {{\cos x - 2\sin x\cos x} \over {2{{\cos }^2}x + \sin x - 1}} = \sqrt 3 \cr & \Leftrightarrow \cos x - 2\sin x\cos x = 2\sqrt 3 {\cos ^2}x + \sqrt 3 \sin x - \sqrt 3 \cr & \Leftrightarrow \cos x - \sin 2x = \sqrt 3 \left( {1 + \cos 2x} \right) + \sqrt 3 \sin x - \sqrt 3 \cr & \Leftrightarrow \cos x - \sqrt 3 \sin x = \sqrt 3 \cos 2x + \sin 2x \cr & \Leftrightarrow {1 \over 2}\cos x - {{\sqrt 3 } \over 2}\sin x = {{\sqrt 3 } \over 2}\cos 2x + {1 \over 2}\sin 2x \cr & \Leftrightarrow \cos x\cos {\pi \over 3} - \sin x\sin {\pi \over 3} = \sin 2x\cos {\pi \over 3} + \cos 2x\sin {\pi \over 3} \cr & \Leftrightarrow \cos \left( {x + {\pi \over 3}} \right) = \sin \left( {2x + {\pi \over 3}} \right) \cr & \Leftrightarrow \cos \left( {x + {\pi \over 3}} \right) = \cos \left( {{\pi \over 6} - 2x} \right) \cr & \Leftrightarrow \left[ \matrix{x + {\pi \over 3} = {\pi \over 6} - 2x + k2\pi \hfill \cr x + {\pi \over 3} = - {\pi \over 6} + 2x + k2\pi \hfill \cr} \right. \Leftrightarrow \left[ \matrix{3x = - {\pi \over 6} + k2\pi \hfill \cr - x = - {\pi \over 2} + k2\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{x = - {\pi \over {18}} + {{k2\pi } \over 3}\,\,\,\left( {tm} \right) \hfill \cr x = {\pi \over 2} - k2\pi \,\,\,\,\,\,\left( {ktm} \right) \hfill \cr} \right.\, \Rightarrow x = - {\pi \over {18}} + {{k2\pi } \over 3}\,\,\left( {k \in Z} \right) \cr} \)

Chọn D.