Take a look at the following problem:
From $x-axis$, we get this equation: $$ -275\cos(\theta)+T_{AC}\cos(30) = 0 \implies T_{AC} = \frac{275}{\cos(30)}\cos(\theta) $$ From $y-axis$, we get this equation:
$$ \begin{align} 275\sin(\theta) + T_{AC}\sin(30) - 300 &= 0 \\ 275\sin(\theta) + \left[ \frac{275}{\cos(30)}\cos(\theta) \right] \sin(30) - 300 &= 0 \\ \tag{1} 275\sin(\theta) + 158.7713\cos(\theta) &= 300 \\ 0.9167\sin(\theta) + 0.5292\cos(\theta) &= 1 \tag{2} \end{align} $$ Solving Eq(2), we get $\theta_{1,2}=40.86678^\circ, 79.13807^\circ$. Which angle should I choose? The only thing I can do is to plug these values in Eq(1) and check if the equation is satisfied. If I do so, I get 0.0013,-0.0075. Obviously the error is a bit large for two values. Note: I'm assuming the figure is not drawn to scale.
The fact that there are two roots to your equation simply means that there are two possible states in which the system can meet the criteria given. Keep in mind that the system doesn't have one but two degrees of freedom (the mass of $C$ and the angle $\theta$). All we end up having physically is that there is one solution where $\theta$ is larger and mass of $C$ is smaller and then there is another solution where $\theta$ is smaller and mass of $C$ is larger. A rigorous answer sheet would note both solutions.