Giải phương trình: sin³ x + cos³ x − sin x − cos x = cos 2x.

Lời giải

Ta có sin³ x + cos³ x − sin x − cos x = cos 2x

Û (sin x + cos x)(sin² x − sin x.cos x + cos² x) − (sin x + cos x) − (cos² x − sin² x) = 0

Û (sin x + cos x)(1 − sin x.cos x) − (sin x + cos x) − (sin x + cos x)(cos x − sin x) = 0

Û (sin x + cos x)(1 − sin x.cos x − 1 − cos x + sin x) = 0

Û (sin x + cos x)(− sin x.cos x − cos x + sin x) = 0

• TH1: sin x + cos x = 0

\( \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\)

\( \Leftrightarrow x = - \frac{\pi }{4} + k\pi \;\left( {k \in \mathbb{Z}} \right)\).

• TH2: − sin x.cos x − cos x + sin x = 0 (1)

Đặt t = sin x − cos x; t Î (−2; 2)

\( \Rightarrow \frac{{{t^2} - 1}}{2} = - \sin x.\cos x\)

Phương trình (1) Û \(t + \frac{{{t^2} - 1}}{2} = 0 \Leftrightarrow {t^2} + 2t - 1 = 0\)

\( \Rightarrow \left[ \begin{array}{l}t = - 1 + \sqrt 2 \;\left( {TM} \right)\\t = - 1 - \sqrt 2 \;\left( {KTM} \right)\end{array} \right.\)

\( \Rightarrow \sin x - \cos x = - 1 + \sqrt 2 \)

\( \Rightarrow \sqrt 2 \cos \left( {x + \frac{\pi }{4}} \right) = - \sqrt 2 + 1\)

\( \Leftrightarrow \cos \left( {x + \frac{\pi }{4}} \right) = \frac{{1 - \sqrt 2 }}{{\sqrt 2 }}\)

\( \Rightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + {\mathop{\rm arc}\nolimits} \,cos\frac{{1 - \sqrt 2 }}{{\sqrt 2 }} + k2\pi \\x = - \frac{\pi }{4} - {\mathop{\rm arc}\nolimits} \,cos\frac{{1 - \sqrt 2 }}{{\sqrt 2 }} + k2\pi \end{array} \right.\;\left( {k \in \mathbb{Z}} \right)\).