Question

Solve on the interval [0,2π):

(sinx+1)(2sin^2x-3sinx-2)= 0

A. x=3π/2, x=7π/6, x=11π/6
B. x=π, x=2π/3, x=5π/3
C. x=2π, x=π/2, x=π/3
D. x=2π, x=π/2, x=5π/4

Answer 1

a

Answer 2

kassia got more?
too much trig for me... ;P