Question

solve for 0=/<x<2pi
6sin^2x + sinx - 1 = 0
Ahh, forgot how to factor.

3cos^2 - 2cos - 1 = 0
(3cosx + 1 )(cosx - 1)
cosx= -1/3, 1
x=0, and .....?

Answer 1

You seem to be doing two different problems.
The first one can be factored to
(3 sinx -1)(2sin x + 1) = 0
sin x = 1/3 or -1/2

x = 19.47, 160.53, 150, 330 (degrees)

Express them in radians if necessary

19.47 degrees is arcsin (1/3) = 0.3398 radians

Answer 2

solve for 0=/<x<2pi
6sin^2x + sinx - 1 = 0
Ahh, forgot how to factor.

6 y^2 + y - 1 = 0
(3 y-1)(2y+1) = 0
y = sin x = 1/3
y = sin x =-1/2

3cos^2 - 2cos - 1 = 0
(3cosx + 1 )(cosx - 1)
cosx= -1/3, 1
x=0, and .....? 2 pi
x = 1.91 radians and 4.37 radians

Answer 3

so x = 0.3398, 2.807, 7pi/6 and 11pi/6

thanks!