Solve the following trigonometric equation where 0° <= x <= 360°. 6cosxsin^2x + cos^2x - 5cosx = 0?

6 cosx sin²x + cos²x - 5cosx = 0
6 cosx (1 - cos²x) + cos²x - 5cosx = 0
6 cosx - 6 cos³x + cos²x - 5cosx = 0
6 cos³x - cos²x - cosx = 0
cosx (6 cos²x - cosx - 1) = 0
cosx (2 cosx - 1) (3cosx + 1) = 0
cosx = 0 or cosx = 1/2 or cosx = -1/3
x = 90°, 270° or x = 60°, (360-60)° or x = (180-70.5)° or (180+70.5)°
x = 60°, 90°, 109.5°, 250.5°, 270°, 300°

NB
Sin^2 x = 1 - Cos^2 x
Substitute in
6Cosx(1 - Cos^2 x) + Cos^2x - 5Cosx = 0
6CosX - 6Cos^3x + Cos^2x - 5Cosx = 0
-6Cos^3x + Cos^2x + Cosx = 0
Factor out 'Cosx'
Hence
CosX ( -6Cos^2x + Cosx + 1) = 0
Hence
CosX = 0
x = Cos^-1(0)
x = 90, 270 ...
&
-6Cos^2x + Cosx + 1) = 0
Apply the Quadratic Eq'n
CosX = {- 1 +/-sqrt[1^2 - 4(-6)(1)]}/ 2(-6)
CosX = { -1 +/- sqrt[1+ 24] } / -12
CosX = { - 1 +/- 5} / -12
CosX = 1/12 + 5/12 = 6/12 = 1/2
CosX = 1/2
X = Cos^-1(1/2)
x = 60 , 300
&
CosX = -1/-12 + 5/12 = 1/12 - 5/12 = -4/12 = -1/3
CosX = (-1/3)
X = Cos^-1(-1/3)
X = 109.47 , 250.53...

Hence
x = 60, 90,109.47, 250.53, 270, & 300.