From the given equation, \[\cos (\pi \sin x) = \sin (\pi \cos x) = \cos \left( \frac{\pi}{2} - \pi \cos x \right).\]This means $ \pi \sin x $ and $ \frac{\pi}{2} - \pi \cos x $ either add up to a multiple of $ 2 \pi, $ or differ by a multiple of $ 2 \pi $. In the first case, \[\pi \sin x + \frac{\pi}{2} - \pi \cos x = 2 \pi n\]for some integer $ n $. Then \[\sin x - \cos x = 2n - \frac{1}{2}.\]Since \[(\sin x - \cos x)^2 = \sin^2 x - 2 \sin x \cos x + \cos^2 x = 1 - \sin 2x \le 2,\]it follows that $ |\sin x - \cos x| \le \sqrt{2} $. Thus, the only possible value of $ n $ is 0, in which case \[\sin x - \cos x = -\frac{1}{2}.\]Squaring, we get \[\sin^2 x - 2 \sin x \cos x + \cos^2 x = \frac{1}{4}.\]Then $ 1 - \sin 2x = \frac{1}{4}, $ so $ \sin 2x = \frac{3}{4} $. In the second case, \[\pi \sin x + \pi \cos x - \frac{\pi}{2} = 2 \pi n\]for some integer $ n $. Then \[\sin x + \cos x = 2n + \frac{1}{2}.\]By the same reasoning as above, the only possible value of $ n $ is 0, in which case \[\sin x + \cos x = \frac{1}{2}.\]Squaring, we get \[\sin^2 x + 2 \sin x \cos x + \cos^2 x = \frac{1}{4}.\]Then $ 1 + \sin 2x = \frac{1}{4}, $ so $ \sin 2x = -\frac{3}{4} $. Thus, the possible values of $ \sin 2x $ are $ \boxed{\frac{3}{4}, -\frac{3}{4}} $.