We know that \begin{align*}(5\cos{\alpha} + 5i\sin{\alpha})^{2004} &= 5^{2004}\cos{(2004\alpha)} + 5^{2004}i\sin{(2004\alpha)}\\ (5\cos{\alpha} - 5i\sin{\alpha})^{2004} &= 5^{2004}\cos{(2004\alpha)} - 5^{2004}i\sin{(2004\alpha)}\end{align*} So \[\begin{eqnarray*}x &=& 5^{2003}\sin{(2004\alpha)}\\ &=& \frac{(5\cos{\alpha} + 5i\sin{\alpha})^{2004} - (5\cos{\alpha} - 5i\sin{\alpha})^{2004}}{10i}\\ &=& \frac{(\pm 4 + 3i)^{2004} - (\pm 4 - 3i)^{2004}}{10i}\] If $\cos{\alpha} = 4/5$, then the $4$ is positive. For the numerator, we only expand out the first two and last two terms because all the others are divisible by 10 and therefore won't be counted in the fractional part. This leaves us with a number congruent to 2 mod 10 as the numerator, so one value is $\boxed{0.2}$. If $\cos{\alpha} = -4/5$, then the $4$ is negative. We do the same thing with $\cos{\alpha} = -4/5$ to get $\boxed{0.8}$ as the other value.