What is cos x if cscx=4squareroot3/3 and x is in the interval (pi;3pi/2)?

Recall that sin(x) =
1/csc(x).

sin(x) = 1/(4sqrt(3)/3) =
3sqrt(3)/[4sqrt(3)*sqrt(3)] = 3sqrt(3)/(4)(3) = 3sqrt(3)/12 =
sqrt(3)/4

Now lets visualize this in a right triangle.  The
sine is opposite over hypotenuse.  So we can say that the opposite is sqrt(3) and the
hypotenuse is 4.

Using the Pythagorean Theorem, we can
solve for the adjacent side.  We need it for the cosine, which is adjacent over
hypotenuse. 

x^2 + sqrt(3)^2 =
4^2

x^2 + 3 = 16

x^2 =
13

x = sqrt(13)

cos(x) would
be adjacent over hypotenuse, so we have:

cos(x) =
sqrt(13)/4 but in Quadrant 3, the interval (pi, 3pi/2), the cosine is
negative.

Therefore, the answer is: 
-sqrt(13)/4