Answer:
    6 < x < 23.206
Step-by-step explanation:
To properly answer this question, we need to make the assumption that angle DAC is non-negative and that angle BCA is acute.
The maximum value of the angle DAC can be shown to occur when points B, C, and D are on a circle centered at A*. When that is the case, the sine of half of angle DAC is equal to 16/22 times the sine of half of angle BAC. That is, ...
   (2x -12)/2 = arcsin(16/22×sin(24°))
   x ≈ 23.206°
Of course, the minimum value of angle DAC is 0°, so the minimum value of x is ...
   2x -12 = 0
   x -6 = 0 . . . . . divide by 2
   x = 6 . . . . . . . add 6
Then the range of values of x will be ...
   6 < x < 23.206
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* One way to do this is to make use of the law of cosines:
   22² = AB² + AC² -2·AB·AC·cos(48°)
   16² = AD² + AC² -2·AD·AC·cos(2x-12)
The trick is to maximize x while satisfying the constraints that all of the lengths are positive. This will happen when AB=AC=AD, in which case the equations be come ...
   22² = 2·AB²·(1-cos(48°))
   16² = 2·AB²·(1 -cos(2x-12))
The value of AB drops out of the ratio of these equations, and the result for x is as above.