How do you solve this trig equation

If -5sinx+2=6cos^2x where x is between 0 and 2π, then one of the factors used to solve for x is
A. 3sinx-4
B. 2sinx-1
C. 6sinx-1
D. sinx+1

I know the answer is supposed to be A, but I got B. can anyone show me how to get the answer? You can explain usung a graphing calculator if you want.

Use the identity cos^2(x) = 1 - sin^2(x):
0 = 6(1 - sin^2(x)) + 5sin(x) - 2
0 = -6sin^2(x) + 5sin(x) + 4
Let u = sin(x)
0 = -6u² + 5u + 4
0 = 6u² - 5u - 4
0 = (3u - 4)(2u + 1)
Reverse the substitution:
0 = (3sin(x) - 4)(2sin(x) + 1)

-5sinx+2=6 - 6sin^2(x)
6sin^2(x)-5sinx-4 = 0
(3sinx-4)(2sinx+1) = 0
Answer A