Asked by lilpump
cosxtanx-sin^2x=0
solve for x over the domain
0 is less than or equal to x which is less than or equal to 2pi
solve for x over the domain
0 is less than or equal to x which is less than or equal to 2pi
Answers
Scott
sin(x) - sin^2(x) = 0
sin(x) [1 - sin(x)] = 0
sin(x) = 0
sin(x) = 1
sin(x) [1 - sin(x)] = 0
sin(x) = 0
sin(x) = 1
lilpump
x= pi, 0, 2pi, and pi/2 ?