Good question. You're right, you can't really throw out one of the solutions. The author of this PDF is using a poorly explained shortcut, which goes like this: if you expand out the full complex solution, you get
$$\begin{align}
\Psi(\theta)
&= c_1 e^{i\omega t} + c_2 e^{-i\omega t} \\
&= (a_1 + b_1 i)(\cos\omega t + i\sin\omega t) + (a_2 + b_2 i)(\cos\omega t - i\sin\omega t) \\
&= (a_1 + a_2)\cos\omega t + (b_2 - b_1)\sin\omega t + i[(a_1 - a_2)\sin\omega t + (b_1 + b_2)\cos\omega t]
\end{align}$$
If you know, for reasons I won't get into here, that you only have to consider the real solution, you can assume "without loss of generality" (as they say) that the imaginary part is zero:
$$\begin{align}a_1 &= a_2 & b_1 &= -b_2\end{align}$$
That allows you to eliminate half the coefficients; in other words, two of $a_1,a_2,b_1,b_2$ are not independent, but are related to the remaining two.
You can now combine the previous two equations to get
$$\Psi(\theta) = 2a_1\cos\omega t - 2b_1\sin\omega t$$
Notice that the solution now has only two real coefficients, $a_1$ and $b_1$, or equivalently, one complex coefficient, $c_1$. The catch is that $c_1$ isn't really being used as a complex number; rather, it's being picked apart and its components used separately, which is why I don't think the way it's explained in the PDF is very useful.