How to compute the charge as a function of time in a simple $CL$ circuit?

I am trying to compute $q(t)$ in a series circuit consisting of a capacitor of plate charges $q_0$ and $-q_0$ and an inductor. The switch is closed at time $t=0$.

By Kirchoff’s Law, $$frac{q}{C}+Lfrac{dI}{dt}=0.$$ There is a conservation of charge in this circuit, so $I=dot{q}$, which gives us the differential equation $$ddot{q}+frac{1}{LC}q=0.$$ I recognise that this is the equation of a harmonic oscillator with angular frequency $frac{1}{sqrt{LC}}$, so $$ddot{q}+omega^2 q=0.$$ To solve this I construct a characteristic equation $$lambda^2+omega^2=0 implies lambda=pm iomega$$ and taking only the positive root gives the solution $q(t)$ as $$q(t)=Acos{omega t}+Bisin{omega t}.$$ The total initial charge in the circuit is from the capacitor and is given by $q(0)=CV$, so we get $Acos{omega t}=CV implies A=CV.$ Then $$q(t)=CVcos(omega t)+Bisin(omega t).$$ I know that the answer should not have the imaginary term in it, but how to I eliminate it or show that B=0?