Since each sine is at most $1,$ we need $\sin(mx) = \sin(nx) = 1,$ i.e. $mx = \frac{\pi}{2} + 2\pi a$ and $nx = \frac{\pi}{2} + 2\pi b$ for integers $a, b.$ Eliminating $x$ gives $n(4a + 1) = m(4b + 1),$ that is, $4(na - mb) = m - n.$ As $a$ and $b$ range over the integers, $na - mb$ takes exactly the multiples of $g = \gcd(m, n),$ so a solution exists if and only if $4g \mid n - m$ — equivalently, writing $m = gm'$ and $n = gn',$ if and only if $m' \equiv n' \pmod 4$ (which forces both $m'$ and $n'$ odd).

For each $g$ we count coprime pairs $m' \lt n' \le \lfloor 30/g \rfloor$ of odd numbers in the same class mod $4.$ For $g = 1:$ among $1, 3, \ldots, 29$ there are eight numbers $\equiv 1$ and seven $\equiv 3 \pmod 4,$ giving $\binom{8}{2} + \binom{7}{2} = 49$ pairs, of which the five pairs $\{3,15\},$ $\{3,27\},$ $\{9,21\},$ $\{15,27\},$ $\{5,25\}$ are not coprime, leaving $44.$ For $g = 2$ (odd numbers up to $15$): $\binom{4}{2} + \binom{4}{2} = 12$ minus the pair $\{3,15\}$ gives $11.$ For $g = 3$ (up to $10$): the pairs $\{1,5\}, \{1,9\}, \{5,9\}, \{3,7\}$ give $4.$ For $g = 4$ (up to $7$): $\{1,5\}$ and $\{3,7\}$ give $2.$ For $g = 5$ and $g = 6:$ only $\{1,5\},$ giving $1$ each. For $g \ge 7$ we would need two distinct odd numbers up to $\lfloor 30/g \rfloor \le 4$ in the same class mod $4,$ which is impossible.

The total is $44 + 11 + 4 + 2 + 1 + 1 = 63.$