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

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1. 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} $$

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