So the given material and equations I have are $$(1):x=C_1\cos(t)+C_2\sin(t)$$ $$x \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$$ $$x'\left(\frac{\pi}{3}\right)=0$$
First I sub in the $\frac{\pi}{3}$ and the $\frac{\sqrt{3}}{2}$ into my first equation and get
$$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{\sqrt{3}}{2}C_2$$
Then finding $x'$and subbing in the same I get $$0=-\frac{\sqrt{3}}{2}C_1+\frac{1}{2}C_2$$ Now I am supposed to add the two equations together to find either $C_1$ or $C_2$ like so $$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{\sqrt{3}}{2}C_2$$ $$0=-\frac{\sqrt{3}}{2}C_1+\frac{1}{2}C_2$$
But neither of the terms cancel out and I'm not sure if i need to find $x''$ to proceed
$$0=-\frac{\sqrt{3}}{2}C_1+\frac{1}{2}C_2$$ $$ \implies \sqrt{3}C_1=C_2$$ Plug that in first equation. $$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{\sqrt{3}}{2}C_2$$ $$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{{3}}{2}C_1$$ $$C_1=\frac{\sqrt{3}}{4}$$ $$ \implies C_2=\sqrt{3}C_1=\frac {3} 4$$