How can I solve this trigonometric equation

How can I solve this trigonometric equation?

6sin^2x-5cosx-2=0

I know that you factor it if it is in quadratic form,but thats for when theyre both the same tri ratio.How can I solve it if the ratios,sin and cos,are both different?Convert one of them using an identity?Thanks in advance.

Use the identity sin²x = 1 – cos²x

6 – 6cos²x – 5cosx – 2 = 0

6cos²x + 5cosx – 4 = (2cosx – 1)(3cosx + 4) = 0

2cosx – 1 = 0 ⇒ cosx = 1/2 ⇒ x = π/3 + 2kπ, 5π/3 + 2kπ where k is any integer
3cosx + 4 = 0 ⇒ cosx = –4/3, no solutions

Ok Remember that sin^2(x) = 1-cos^2(x)
So you have 6*(1-cos^2(x)) - 5*cos(x) -2 = 0.
Now distribute to get 6 -6cos^2(x) -5*cos(x) -2 = 0.
Now, add the constants and multiply thru by -1 to get
6cos^2(x) +5*cos(x) -4 = 0
This factors to (3*cos(x) +4)*(2cos(x)-1) = 0
Now by ZPP cos(x) = -4/3 which cannot be true or cos(x) = 1/2 which is true in Q1 and Q4
x = pi/3 (60 degrees) and 300 degrees or 10pi/6.
Now, I don't know how picky your teacher is but if you add multiplies of 2pi to each of these you'll get a bunch of answers. Originally, just dealing with the 0-2pi interval.

6(1-cos^2 x) - 5cosx - 2 = 0 , simplify and solve