What is the domain of f(t)=ln(sin(t))

The domain of ln(x) is (0, infinity).
so we need to solve sin(t) > 0

sin is positive in the first and second quadrants,

so the domain is

...(-4π, -3π) U (-2π, -π) U (0, π) U (2π, 3π) U (4π, 5π) ...

ln is defined for all positive reals. sin(t) is positive for floor((t-pi)/pi) an odd number, when t is not a multiple of pi. Then we can define the domain recursively.

The domain of f(t), for -2*n*pi<=t<=2*n*pi, is b_n, where

b_0 = (0,pi)

For n>0,

b_(n+1) = b_n u (2*(n+1)*pi, (2*(n+1) +1)*pi) u (-2*(n+1)*pi, (-2*(n+1) +1)*pi)

"u" denotes set union and (x,y) denotes the open interval from x to y on the real number line.