Find all the solutions in the interval [0, 2pi): 2 sin^3 x + sin^2 x = 0.?

Question

a. 5pi/6, 11pi/6
b. 4pi/3, 5pi/3
c. 0, 7pi/6, pi, 11pi/6
d. 0, pi/2, pi, 4pi/3, 3pi/2, 5pi/3
e. none of these

Answer 1

sinx^2(2sinx+1)=0 so
sinx^2=0 sinx=0=>x=0; and x=pi
and 2sinx+1=0=>sinx=-1/2 =>x=2pi-pi/6=>x=11pi/6
x=pi+pi/6=7pi/6
so C

Answer 2

sin²x(2sinx + 1) = 0
sin x = 0, sinx = - 1/2
x= 0, 7π / 6 and 11π / 6
Answer c