if a language $L \in$ coNP, i.e. it's complement is in NP, then does L have a deterministic turing machine that decides it?
i think that this is false, but am unsure how to show it? my guess is using the fact that L is in NP, but i thought problems in NP do not necessarily need a deterministic turing machine to decide it, just a deterministic turing machine to verify it
Actually because $NP$ and $coNP$ are subsets of $EXP$, there always is a deterministic touring machine deciding any of these. The problem is that it is a machine with a very long running time.
For an $NP$-language $L$, a deterministic $EXP$-TM deciding $L$ is simply the "try all verification candidates" (These are of polynomially bounded length, also called (nondeterministic) computation paths of the $NP$-TM) in order. If one of these paths was accepting, accept, else decline.
An analogous construction is possible for $L\in coNP$.