Ta có:
$4(x^3-2x+1)(\sin x+2\cos x)\ge9|x^3-2x+1|$
$\Rightarrow 16(\sin x+2\cos x)^2(x^3-2x+1)^2\ge81(x^2-2x+1)^2$
$\Leftrightarrow (x^3-2x+1)^2[16(\sin x+2\cos x)^2-81]\ge0 (*)$
Mà ta có:
$(\sin x+2\cos x)^2\le (1^2+2^2)(\sin^2x+\cos^2x)=5$
$\Rightarrow 16(\sin x+2\cos x)^2-81<0$
Từ đó: $(*) \Leftrightarrow x^3-2x+1=0 \Leftrightarrow \left[\begin{array}{l}x=1\\x=\dfrac{-1\pm\sqrt5}{2}\end{array}\right.$, thỏa mãn.