$f(x) =$ cos3x−cosx+cos2x−1
$=4\cos^3 x -3\cos x -\cos x +2\cos^2 x - 1 -1 = 4\cos^3 x +2\cos^2 x -4\cos x - 2$
$=2\cos^3 x +\cos^2 x -2\cos x-1$
đặt $\cos x= t ;\ t\in [-1;\ 1] \Rightarrow f(t) = 2t^3 +t^2 -2t-1$
$f'(t) =6t^2 +2t -2;\ f'(t) =0 \Rightarrow t= \dfrac{1}{6}(-1 \pm \sqrt{13})$
bạn tính $f(\pm 1);\ f(\dfrac{1}{6}(-1 \pm \sqrt{13}))$ rồi so sánh đưa ra min max nhé