Solve 6 sin^(2)x - 5 cos x - 2 = 0 in the interval 0 ≤ x ≤ 2pi

Answer 1

since sin^2x = 1-cos^2x, that means
6(1-cos^2x) - 5cosx - 2 = 0
6 - 6cos^2x - 5cosx - 2 = 0
6cos^2x + 5cosx - 4 = 0
(3cosx+4)(2cosx-1) = 0
so, where do you have
cosx = -4/3 ? nowhere
where is cosx = 1/2 ?