Since $ 135^\circ < x < 180^\circ, $ $ \cos x < 0 < \sin x $ and $ |\sin x| < |\cos x| $. Then $ \tan x < 0, $ $ \cot x < 0, $ and \[|\tan x| = \frac{|\sin x|}{|\cos x|} < 1 < \frac{|\cos x|}{|\sin x|} = |\cot x|.\]Therefore, $ \cot x < \tan x $. Furthermore, $ \cot x = \frac{\cos x}{\sin x} < \cos x $. This tells us that for the four points $ P, $ $ Q, $ $ R, $ $ S $ that lie on the parabola $ y = x^2, $ $ P $ and $ S $ are between $ Q $ and $ R $. Hence, the parallel bases of the trapezoid must be $ \overline{PS} $ and $ \overline{QR} $. Then their slopes must be equal, so \[\cos x + \tan x = \cot x + \sin x.\]Then \[\cos x + \frac{\sin x}{\cos x} = \frac{\cos x}{\sin x} + \sin x,\]so \[\cos^2 x \sin x + \sin^2 x = \cos^2 x + \cos x \sin^2 x.\]Then $ \cos^2 x \sin x - \cos x \sin^2 x + \sin^2 x - \cos^2 x = 0, $ which we can factor as \[(\sin x - \cos x)(\cos x + \sin x - \sin x \cos x) = 0.\]Since $ \cos x < 0 < \sin x, $ we must have \[\cos x + \sin x = \sin x \cos x.\]We can write this as \[\cos x + \sin x = \frac{1}{2} \sin 2x.\]Squaring both sides, we get \[\cos^2 x + 2 \sin x \cos x + \sin^2 x = \frac{1}{4} \sin^2 2x,\]so $ \sin 2x + 1 = \frac{1}{4} \sin^2 2x, $ or $ \sin^2 2x - 4 \sin 2x - 4 = 0 $. By the quadratic formula, \[\sin 2x = 2 \pm 2 \sqrt{2}.\]Since $ -1 \le \sin 2x \le 1, $ we must have $ \sin 2x = \boxed{2 - 2 \sqrt{2}} $.