Question

Solve on the interval [0,2Pi):
(sinx + 1) (2sin^2x-3sinx-2)= 0

A. x =2Pi, x =pi/2, x =5pi/4
B. x =3pi/2, x =7pi/6, x =11pi/6
C. x =2pi, x =pi/2, x =pi/3
D. x =pi, x =2pi/3, x =5pi/3
Answer with an explaination?

Answer 1

it was b...

Answer 2

you delete all of your explaination and your answer so i just look retarded.... goodjob