If $cos^{-1} (cos 4) > 3x^2 - 4x$, then

$x ∈ \left( -∞,\frac{2-\sqrt{6\pi - 8}}{3}\right)$ $x ∈ \left( \frac{2+\sqrt{6\pi - 8}}{3}, ∞\right)$ $x ∈ \left( \frac{2-\sqrt{6\pi - 8}}{3}, \frac{2+\sqrt{6\pi - 8}}{3}\right)$ ∞

$x ∈ \left( \frac{2-\sqrt{6\pi - 8}}{3}, \frac{2+\sqrt{6\pi - 8}}{3}\right)$

We have  $cos^{-1}(cos4) > 3x^2 - 4x $ $⇒cos^{-1}(cos(2\pi-4)) > 3x^2 - 4x $ $⇒ 2 \pi - 4 > 3x^2 - 4x $ $⇒3x^2 - 4x -(2 \pi - 4) < 0$ $⇒\frac{2-\sqrt{6\pi - 8}}{3}< x< \frac{2+\sqrt{6\pi - 8}}{3}$