$9-12\cos x+4{\cos }^{2}x=-8\left({\cos }^2\left(\frac{x}{2}\right)-{\sin }^2\left(\frac{x}{2}\right)\right)+12$
$9-12\cos x+4{\cos }^{2}x=-8\cdot \cos \left(2\cdot \frac{x}{2}\right)+12$
$9-12\cos x+4{\cos }^{2}x=-8\cos x+12$
$4{\cos }^{2}x-4\cos x-3=0$

$t=\cos x,t\in \left[-1,1\right]$ $4t^2-4t-3=0$ $\Delta ={\left(-4\right)}^2-4\cdot 4\cdot \left(-3\right)=64,\sqrt{\Delta }=8$ $t_1=\frac{4-8}{2\cdot 4}=-\frac{4}{8}=-\frac{1}{2}$ $t_2=\frac{4+8}{2\cdot 4}=\frac{12}{8}\notin \left[-1,1\right]$

$\cos x=-\frac{1}{2}$ $x\in \left(0,\pi \right)$, czyli: $x=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$

Odp.: $x=\frac{2\pi }{3}$