Recall that
sin(x) = (e^ix - e^-ix)/2i
cos(x) = (e^ix + e^-ix)/2
3 sin(4x) = 3(e^4ix - e^-4ix)/2i
4 cos(10x) = 4(e^10x + e^-10x)/2
Simplify as you will. Not sure what you mean by "functional series" involving e^ix. Using those definitions, there is no series involved.
If you just want series expansions, recall that
sin(x) = x - x^3/3! + x^5/5! - ...
cos(x) = 1 - x^2/2! + x^4/4! - ...
make substitutions for your values, and rewrite as needed.
expand 3sin(4x)+4cos(10x) in to functional series of e(exp)ix
2 answers
that should be
4 cos(10x) = 4(e^10ix + e^-10ix)/2
4 cos(10x) = 4(e^10ix + e^-10ix)/2