The set of all values of x, measured in radians, satisfying the two inequalities $2 \cos^2 x <2-3 \cos x$ and $x^2 <4x+12$, is

$(-2, -π/3)∪(π/3, 5π/3)$

We have,

$2 \cos^2 x <2-3 \cos x$ and $x^2 <4x+12$

Now,

$2 \cos^2 x <2-3 \cos x$

$⇒2 \cos^2 x + 3 \cos x-2<0$

$⇒(2 \cos x-1) (\cos x + 2) <0$

$⇒2 \cos x-1 <0⇒ \cos x<\frac{1}{2}$   $[∵\cos x+2>0]$   ...(i)

and,

$x^2 <4x + 12$

$⇒x^2-4x-12<0⇒(x-6) (x+2) <0⇒>-2<x<6$  ...(ii)

Now, we have to find the values of x which satisfy in equations (i) and (ii).

We know that $-π <-2<-\frac{π}{2}<\frac{π}{2}<2<π<\frac{3π}{2}<6<2π$

$∴\cos x<\frac{1}{2}$ if $x∈(-2, -π/3)∪(π/3, 5π/3)$

Hence, the set of values of x satisfying the given inequations is $(-2, -π/3)∪(π/3, 5π/3)$