$\sin x\le\cos x$ is equivalent with $x\in [2k\pi,\pi/4+2k\pi]\cup [5\pi/4+2k\pi,2\pi+2k\pi )$ $1-\sinh\le 1-\cos x$ is equivalent with $x\in [\pi/4+2k\pi,5\pi/5+2k\pi]$ cases: 1) $x\in [2k\pi,\pi/4+2k\pi]$ $\sin x<1-\cos x$ , no solution! 2) $x\in (\pi/4+2k\pi,5\pi/4+2k\pi )$ $\cos x<1-\sin x$ , solution is $x\in (\pi/2+2k\pi,5\pi/4+2k\pi )$ 3) $x\in [5\pi/4+2k\pi,2\pi+2k\pi )$ $\sin x<1-\cos x$ , solution is $x\in [5\pi/4+2k\pi,2\pi+2k\pi )$ finally $\boxed{x\in (\pi/2+2k\pi,2\pi+2k\pi )}$