Giải phương trình đối xứng gần đối xứng sin^3x + cos^3x= 1 + (√2 - 2) sinxcosx 2sin2x - 3√6 |sinx+cosx|+8 = 0

Đáp án:

\(\eqalign{ & 1)\,\,\,\left[ \matrix{ x = k2\pi \hfill \cr x = {\pi \over 2} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr & 2)\,\,\left[ \matrix{ x = {\pi \over {12}} + k2\pi \hfill \cr x = {{5\pi } \over {12}} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} \)

Giải thích các bước giải:

$$\eqalign{ & 1)\,\,{\sin ^3}x + {\cos ^3}x = 1 + \left( {\sqrt 2 - 2} \right)\sin x\cos x \cr & \Leftrightarrow {\left( {\sin x + \cos x} \right)^3} - 3\sin x\cos x\left( {\sin x + \cos x} \right) = 1 + \left( {\sqrt 2 - 2} \right)\sin x\cos x \cr & Dat\,\,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 = {{{t^2} - 1} \over 2} \cr & \Rightarrow {t^3} - 3{{{t^2} - 1} \over 2}t = 1 + \left( {\sqrt 2 - 2} \right){{{t^2} - 1} \over 2} \cr & \Leftrightarrow 2{t^3} - 3{t^3} + 3t = 2 + \left( {2\sqrt 2 - 4} \right){t^2} - \left( {2\sqrt 2 - 4} \right) \cr & \Leftrightarrow {t^3} + \left( {2\sqrt 2 - 4} \right){t^2} - 3t + 6 - 2\sqrt 2 = 0 \cr & \Leftrightarrow \left( {t - 1} \right)\left[ {{t^2} + \left( {2\sqrt 2 - 3} \right)t - 6 + 2\sqrt 2 } \right] = 0 \cr & \Leftrightarrow \left[ \matrix{ t = 1 \hfill \cr t \approx 1,86\,\,\left( {ktm} \right) \hfill \cr t \approx - 1,7\,\,\left( {ktm} \right) \hfill \cr} \right. \Leftrightarrow t = 1 \cr & \sin x + \cos x = 1 \Leftrightarrow \sin \left( {x + {\pi \over 4}} \right) = {1 \over {\sqrt 2 }} \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 \over 2} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr & 2)\,\,2\sin 2x - 3\sqrt 6 \left| {\sin x + \cos x} \right| + 8 = 0 \cr & t = \left| {\sin x + \cos x} \right|\,\,\left( {0 \le t \le \sqrt 2 } \right) \cr & \Rightarrow {t^2} = 1 + 2\sin x\cos x \Leftrightarrow \sin 2x = {t^2} - 1 \cr & 2\left( {{t^2} - 1} \right) - 3\sqrt 6 t + 8 = 0 \cr & \Leftrightarrow 2{t^2} - 3\sqrt 6 t + 6 = 0 \cr & \Leftrightarrow \left[ \matrix{ t = \sqrt 6 \,\,\left( {ktm} \right) \hfill \cr t = {{\sqrt 6 } \over 2}\,\,\left( {tm} \right) \hfill \cr} \right. \cr & \Rightarrow \sin x + \cos x = {{\sqrt 6 } \over 2} \cr & \Leftrightarrow \sqrt 2 \sin \left( {x + {\pi \over 4}} \right) = {{\sqrt 6 } \over 2} \cr & \Leftrightarrow \sin \left( {x + {\pi \over 4}} \right) = {{\sqrt 3 } \over 2} \cr & \Leftrightarrow \left[ \matrix{ x + {\pi \over 4} = {\pi \over 3} + k2\pi \hfill \cr x + {\pi \over 4} = {{2\pi } \over 3} + k2\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = {\pi \over {12}} + k2\pi \hfill \cr x = {{5\pi } \over {12}} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$