Your equation is linear in c
but it is trascendental in k
. You get a closed-form result for example if you solve first for c
in one of the equation, replace into the other, and give specific values for the parameters:
Solve[exp2 /.
Solve[exp1, c] /. {\[Mu] -> 4, \[Gamma] -> 0, \[Alpha] ->
1, \[Beta] -> 1, \[Nu] -> 1}, x]
It is not clear to me what you mean with the third argument in Solve[{exp1,exp2},{c},{k}]