Giải phương trình lượng giác 2√2sin(x-π/4) - sin2x - 2 =0 2cos5x.cos2x-√3sin7x = sin(5π/2 -3x)+1 Mọi ng giúp mình với ạ :(((

Đáp án:

$\eqalign{ & 1)\,\,\left\{ \matrix{ x = k2\pi \hfill \cr x = \pi + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr & 2)\,\,\left\{ \matrix{ x = {{k2\pi } \over 7} \hfill \cr x = - {{2\pi } \over {21}} + {{k2\pi } \over 7} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $

Giải thích các bước giải:

$\eqalign{ & 1)\,\,\,2\sqrt 2 \sin \left( {x - {\pi \over 4}} \right) - \sin 2x - 2 = 0 \cr & \Leftrightarrow 2\left( {\sin x - \cos x} \right) - 2\sin x\cos x - 2 = 0 \cr & \Leftrightarrow \sin x - \cos x - \sin x\cos x - 1 = 0 \cr & \text{Đặt }t = \sin x - \cos x\,\,\left( { - \sqrt 2 \le t \le \sqrt 2 } \right) \cr & \Rightarrow {t^2} = 1 - 2\sin x\cos x \Leftrightarrow \sin x\cos x = {{1 - {t^2}} \over 2} \cr & \text{Pt: }t - {{1 - {t^2}} \over 2} - 1 = 0 \cr & \Leftrightarrow 2t - 1 + {t^2} - 2 = 0 \cr & \Leftrightarrow {t^2} + 2t - 3 = 0 \Leftrightarrow \left[ \matrix{ t = 1\,\,\left( {tm} \right) \hfill \cr t = - 3\,\,\left( {ktm} \right) \hfill \cr} \right. \cr & t = 1 \Leftrightarrow \sqrt 2 \sin \left( {x - {\pi \over 4}} \right) = 1 \cr & \Leftrightarrow \left[ \matrix{ x - {\pi \over 4} = {\pi \over 4} + k2\pi \hfill \cr x - {\pi \over 4} = {{3\pi } \over 4} + k2\pi \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = k2\pi \hfill \cr x = \pi + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr & 2)\,\,2\,\cos 5x\cos 2x - \sqrt 3 \sin 7x = \sin \left( {{{5\pi } \over 2} - 3x} \right) + 1 \cr & \Leftrightarrow \left( {\cos 7x + \cos 3x} \right) - \sqrt 3 \sin 7x = \sin \left( {2\pi + {\pi \over 2} - 3x} \right) + 1 \cr & \Leftrightarrow \cos 7x + \cos 3x - \sqrt 3 \sin 7x = \cos 3x + 1 \cr & \Leftrightarrow \cos 7x - \sqrt 3 \sin 7x = 1 \cr & \Leftrightarrow {1 \over 2}\cos 7x - {{\sqrt 3 } \over 2}\sin 7x = {1 \over 2} \cr & \Leftrightarrow \cos \left( {7x + {\pi \over 3}} \right) = {1 \over 2} \cr & \Leftrightarrow \left[ \matrix{ 7x + {\pi \over 3} = {\pi \over 3} + k2\pi \hfill \cr 7x + {\pi \over 3} = - {\pi \over 3} + k2\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ 7x = k2\pi \hfill \cr 7x = - {{2\pi } \over 3} + k2\pi \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = {{k2\pi } \over 7} \hfill \cr x = - {{2\pi } \over {21}} + {{k2\pi } \over 7} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $