Solve 6sin^2 x-5cosx-2=0 in the interval 0≤x≤2π

Answer 1

6sin^2 x-5cosx-2=0
6(1 - cos^2 x) - 5cosx - 2 = 0
6 - 6cos^2 x - 5cosx - 2 = 0
6cos^2 x + 5cosx -4 = 0
(2cosx - 1)(3cosx + 4) = 0
cosx = 1/2 or cosx = -4/3, but the latter is not possible since -1 ≤ cosx < +1
x = 60° or x = 300°

in radians: x = π/3 or 5π/3

Answer 2

How to get the x=60° or x=300°? I didn't catch that point.