What values can the expression $x+y+z$ take if $\sin x=\cos y, \sin y=\cos z, \sin z=\cos x, 0 \leq x, y, z \leq$ $\pi / 2 ?$
$\sin x=\sin (\pi / 2-y), \sin y=\sin (\pi / 2-z), \sin z=\sin (\pi / 2-x)$. From the condition, it follows that all expressions under the sine sign are in the first quadrant. Therefore, $x=\pi / 2-y, y=\pi / 2-z, z=\pi / 2-x$. Adding them up, we get $x+y+z=3 \pi / 2-(x+y+z)$, that is, $x+y+z=3 \pi / 4$.

## Answer

$3 \pi / 4$.