Question 822038
solve for x-values between [0,2pi)
6sin^2x+9sinx+3=0
divide by 3
2sin^2x+3sinx+1=0
factor:
(2sinx+1)(sinx+1)=0
..
2sinx+1=0
sinx=-1/2
x=5π/4,7π/4 (in quadrant III and IV in which sin>0)
or
sinx+1=0
sinx=-1
x=3π/2