2 cot^2t sin t - cot^2t = 0

2 cot²(t ) sin(t ) - cot²(t ) = 0

cot²(t ) (2 sin(t ) - 1) = 0

cot²(t ) = 0   or   2 sin(t ) - 1 = 0

cot(t ) = 0   or   sin(t ) = 1/2

cos(t ) / sin(t ) = 0   or   sin(t ) = 1/2

cos(t ) = 0   or   sin(t ) = 1/2

[t = cos⁻¹(0) + 2nπ   or   t = cos⁻¹(0) - π + 2nπ]

  or   [t = sin⁻¹(1/2) + 2nπ   or   t = π - sin⁻¹(1/2) + 2nπ]

(where n is any integer)

t = π/2 + 2nπ   or   t = -π/2 + 2nπ   or   t = π/6 + 2nπ   or   t = 5π/6 + 2nπ

Note that the first two families of solutions overlap and can be condensed, so that

t = π/2 + nπ   or   t = π/6 + 2nπ   or   t = 5π/6 + 2nπ

i’m sorry but i don’t know sorry

she will sew have sewn five dresses   and be 1/3 done with the sixth one in an hour.

step-by-step explanation:

3/4 of an hour=45 mins

lucy makes 1 dress per 11.25 minutes

45 mins+11.25 mins=56.25 mins

60 mins- 56.25 mins=3.75 mins

3.75 mins is 1/3 of of 11.25 mins

therefore she the other 1/3 would be part of the hour and she will have sewn 5 dresses and begun the sixth.

contact me if you have any questions you are welcome.