Giải phương trình: cos3x – 2sin2x – cosx – sinx – 1 = 0.

cos3x – 2sin2x – cosx – sinx – 1 = 0

Û 4cos3x – 3cosx – 4sinx.cosx – sinx – cosx – 1 = 0

Û 4(1 – sin2 x)cosx – 4cosx(sinx + 1) – (sinx + 1) = 0

Û (1 + sinx)[4(1 – sinx)cosx – 4cosx – 1] = 0

Û (1 + sinx)( – 4sinx.cosx – 1) = 0

$\iff \left[\begin{array}{l}\sin x=-1\\ \sin x=-\frac{1}{2}\end{array}\right.\iff \left[\begin{array}{l}x=-\frac{\pi }{2}+k2\pi \\ x=-\frac{\pi }{12}+k\pi \\ x=\frac{7\pi }{12}+k\pi \end{array}\right.(k\in \mathbb{Z})$

Vậy $x=-\frac{\pi }{2}+k2\pi$; $-\frac{\pi }{12}+k\pi$ hoặc $\frac{7\pi }{12}+k\pi$ $(k\in \mathbb{Z})$.