jmc

We have $\left|\frac{1}{2}-ci\right|=\sqrt{{\frac{1}{2}}^2+(-c{)}^2}=\sqrt{c^2+\frac{1}{4}}$, so $\left|\frac{1}{2}-ci\right|=\frac{3}{4}$ gives us $\sqrt{c^2+\frac{1}{4}}=\frac{3}{4}$. Squaring both sides gives $c^2+\frac{1}{4}=\frac{9}{16}$, so $c^2=\frac{5}{16}$. Taking the square root of both sides gives $c=\frac{\sqrt{5}}{4}$ and $c=-\frac{\sqrt{5}}{4}$ as solutions, so there are $\boxed{2}$ real values of $c$ that satisfy the equation.

We also could have solved this equation by noting that $\left|\frac{1}{2}-ci\right|=\frac{3}{4}$ means that the complex number $\frac{1}{2}-ci$ is $\frac{3}{4}$ units from the origin in the complex plane. Therefore, it is on the circle centered at the origin with radius $\frac{3}{4}$. The complex number $\frac{1}{2}-ci$ is also on the vertical line that intersects the real axis at $\frac{1}{2}$, which is inside the aforementioned circle. Since this line goes inside the circle, it must intersect the circle at $\boxed{2}$ points, which correspond to the values of $c$ that satisfy the original equation.