Giải phương trình: cos³x + cos²x + 2sinx – 2 = 0.

cos³x + cos²x + 2sinx – 2 = 0

⇔ cos²x(cosx + 1) + 2(sinx – 1) = 0

⇔ (1 – sinx)(1 + sinx)(cosx + 1) + 2(sinx – 1) = 0

⇔ (1 – sinx)[(1 + sinx)(cosx + 1) – 2] = 0

⇔ (1 – sinx)(1 + sinx + cosx + sinx cosx - 2) = 0

⇔ (1 – sinx)(sinx + cosx + sinx cosx - 1) = 0 (*)

Đặt t = cosx + sinx ($-\sqrt{2}\le t\le \sqrt{2}$ )

2 sinx cosx = t² – 1 ⇔ sinx cosx = $\frac{t^2-1}{2}$

Phương trình (*) trở thành:

(1 – sinx)$\left(t+\frac{t^2-1}{2}-1\right)$ = 0

⇔ (1 – sinx)$\frac{t^2+2t-3}{2}$ = 0

⇔ $\left[\begin{array}{l}\sin x=1\\ t=1\\ t=-3\left(L\right)\end{array}\right.$

⇔ $\left[\begin{array}{l}x=\frac{\pi }{2}+k2\pi \\ \sqrt{2}\sin \left(x+\frac{\pi }{4}\right)=1\end{array}\right.$

⇔  $\left[\begin{array}{l}x=\frac{\pi }{2}+k2\pi \\ \sin \left(x+\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\end{array}\right.$

⇔ $\left[\begin{array}{l}x=\frac{\pi }{2}+k2\pi \\ x+\frac{\pi }{4}=\frac{\pi }{4}+k2\pi \\ x+\frac{\pi }{4}=\frac{3\pi }{4}+k2\pi \end{array}\right.$

⇔ $\left[\begin{array}{l}x=\frac{\pi }{2}+k2\pi \\ x=k2\pi \end{array}\right.\left(k\in \mathbb{Z}\right)$ .