For $135^\circ < x < 180^\circ$, points $P=(\cos x, \cos^2 x), Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)$ and $S =(\tan x, \tan^2 x)$ are the vertices of a trapezoid. What is $\sin 2x$?
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 $\sin 2x = \boxed{2 - 2 \sqrt{2}}.$