Here is a webpage .......look at Hal Angle formulas and see what you find.
Remember Sin and COs are both negative in Quadrant III
sin = -4/5 cos = - 3/5 (because cos^2 = 1- sin^2)
Then go from there ......
I'll use the first evalution to see if I get the same answer as you did:
Tan x/2 = +- sqrt ( (1-cosx) /(1+cos x) )
=+-sqrt ( (1- -3/5)/(1+ -3/5) ) = + - sqrt( 8/8 / 2/5) = +-sqrt (4) =+2 or -2 We know it is the negative one because it is in Q II
So YES! The answer is -2 !!!