Giải phương trình: sin^3 x + cos^3 x − sin x − cos x = cos 2x

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

Trả lời

Hướng dẫn 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

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

$\iff x=-\frac{\pi }{4}+k\pi \left(k\in \mathbb{Z}\right)$

• 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 $t+\frac{t^2-1}{2}=0\iff t^2+2t-1=0$

$$\begin{gathered} \Rightarrow \left[\begin{array}{l}t=-1+\sqrt{2}\left(TM\right)\\ t=-1-\sqrt{2}\left(KTM\right)\end{array}\right. \\ \Rightarrow \sin x-\cos x=-1+\sqrt{2} \\ \Rightarrow \sqrt{2}\cos \left(x+\frac{\pi }{4}\right)=-\sqrt{2}+1 \\ \iff \cos \left(x+\frac{\pi }{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}} \\ \Rightarrow \left[\begin{array}{l}x=-\frac{\pi }{4}+\mathrm{arc}cos\frac{1-\sqrt{2}}{\sqrt{2}}+k2\pi \\ x=-\frac{\pi }{4}-\mathrm{arc}cos\frac{1-\sqrt{2}}{\sqrt{2}}+k2\pi \end{array}\right.\left(k\in \mathbb{Z}\right) \end{gathered}$$