Question -30

We must have either $\cos(x) = \sin(y) = 1$ or $\cos(x) = \sin(y) = -1.$

In the first case there are two values of $x$ on the given interval for which $\cos(x) = 1$ , namely $0$ and $2\pi$ , and two values of $y$ for which $\sin(y) = 1$ , namely $\frac{\pi}{2}$ and $\frac{5\pi}{2}.$ This gives us $2*2 = 4$ ordered pairs.

In the second case there are two values of $x$ on the given interval for which $\cos(x) = -1$ , namely $\pi$ and $3\pi$ , and just one value of $y$ for which $\sin(y) = -1$ , namely $\frac{3\pi}{2}.$ This gives us $2*1 = 2$ ordered pairs.

Thus the total number of ordered solution pairs is $4 + 2 = \boxed{6}.$