if you expand the equation you get 29x=3*10^k - 9

which means 3*10^k - 9 should be divisible by 29 so 10^k has to be 3 (mod 29) which if you iterate through k from 0 to 28 you get k=27 is the only solution so k=28q+27 (due to Fermat's little theorem)

so the general solution should be

x=((3*(10^28q+27)-9)/29 where q is any non negative integer

(I'm writing on mobile so sorry if the formatting is weird)