Sinx + cosx= cosx/1-sin2x

Đáp án:

\(\left[ \matrix{ x = k\pi \hfill \cr x = - {\pi \over 8} + {{k\pi } \over 2} \hfill \cr} \right.\,\,\left( {tm} \right)\)

Giải thích các bước giải:

\(\eqalign{ & \sin x + \cos x = {{\cos x} \over {1 - \sin 2x}}\,\,\left( {x \ne {\pi \over 4} + k\pi } \right) \cr & \Leftrightarrow \sin x + \cos x = {{\cos x} \over {1 - 2\sin x\cos x}} \cr & \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {1 - 2\sin x\cos x} \right) = \cos x \cr & \Leftrightarrow \sin x + \cos x - 2{\sin ^2}x\cos x - 2\sin x{\cos ^2}x = \cos x \cr & \Leftrightarrow \sin x - 2{\sin ^2}x\cos x - 2\sin x{\cos ^2}x = 0 \cr & \Leftrightarrow \sin x\left( {1 - 2\sin x\cos x - 2{{\cos }^2}x} \right) = 0 \cr & \Leftrightarrow \sin x\left( { - \cos 2x - \sin 2x} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ \sin x = 0 \hfill \cr \sin 2x = - \cos 2x \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ \sin x = 0 \hfill \cr \tan 2x = - 1 \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = k\pi \hfill \cr 2x = - {\pi \over 4} + k\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = k\pi \hfill \cr x = - {\pi \over 8} + {{k\pi } \over 2} \hfill \cr} \right.\,\,\left( {tm} \right) \cr} \)