Let $C$ be a smooth projective curve over $\mathbb{C}$. We know $C$ is hyperelliptic if and only if it has a linear system of dimension 1 and degree 2, that is it has $g^{1}_{2}$. $C$ is trigonal if and only if it has a linear system of dimension 1 and degree 3, that is it has $g^{1}_{3}$.
I can prove that if $g(C)\geq 3$, then $C$ can not be both hyperelliptic and trigonal.
However, on the other hand, I get the following. Suppose $C$ is hyperelliptic, then $C$ has $g^{1}_{2}$. So there exists an effective divisor $D$ on $C$ such that $\mathrm{deg}D=2$ and $\mathrm{dim}|D|\geq1$.(not necassarily equal 1 since we could choose the linear subspace). Now choose any point $P$ on $C$, then $\mathrm{deg} (D+P)=3$ and $\mathrm{dim}|D+P|\geq\mathrm{dim}|D|\geq1$. Thus we could get a $g^{1}_{3}$. Thus $C$ is trigonal. However, this is impossible.
Could you tell me where I made a mistake?