SOLUTION: how do i solve the following equation for values of x in the range 0 degrees < x <360 degrees 6sin^2 x - sinx - 2 = 0

6sin²x - sinx - 2 = 0

Let w = sinx

Then w² = sin²x

6w² - w - 2 = 0

Factor as

(2w + 1)(3w - 2) = 0

2w + 1 = 0;   3w - 2 = 0
    2w = -1;      3w = 2
     w = ;      w = 

Now replace w by sinx

   sinx = ;   sinx = 

The first one can be solved as special angles.
The sine is negative in the 3rd and 4th quadrants,
and since the reference angle 30° has sine +,
the angles in the 3rd and 4th quadrants having that
sine are 210° and 330°.

The second one can't be solved as a special angles.
The sine is positive in the 1st and 2nd quadrants,
We find the refrence angle using inverse sine on a
calculator as 41.8103149°. So the angles in the 1st
and 2nd quadrants having that sine are 41.8103149°
for the 1st quadrant answer and to find the 2nd 
quadrant answer we subtract 41.8103149° from 180°.

180° - 41.8103149° = 138.1896861°

The four solutions are:

x =  210°, 330°, 41.8103149°, and 138.1896861°

Edwin