Can someone explain to me why are we adding 2kpi when we are doing zeros for sin and cos, but adding kpi when doing zeros for tg?

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.

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐