on the first exercise, you got a solution angle of π/18, that's a good solution for the I Quadrant only, however, on a circle, we have angles that go from 0 to 2π, however we can always keep on going around and continute to 2π + π/2 or 3π or 4π, or 115π/3 or 1,000,000π/18 and so on, and we're really just going around the circle many times over, getting a larger and larger angle, same circular motion.
π/18 on that exercise works for the I Quadrant, however if we continue and go around say 2π, we'll find that 2π/3 + π/18 is a coterminal angle with π/18, and thus that angle has also the same sine value.
π/18 + 2kπ/3 , where k = integer, is a way to say, all angles around the circle that look like this have the same sine, namely
π/18 + 2(1)π/3
π/18 + 2(2)π/3
π/18 + 2(3)π/3
π/18 + 2(5)π/3
π/18 + 2(99999999)π/3
.....
so using the "k" as some sequence multiplier, is a generic notational way to say, "all these angles".
you'll also find that "n" is used as well for the same notation, say for example
2π/3 + 2πn.