Let y=cos(3x) then:
4y3-12y2+9y-2=0.
When y=2, this is 4×8-12×4+9×2-2=32-48+18-2=0 so 2 is a zero. Divide by this zero:
2 | 4 -12 9 -2
4 8 -8 | 2
4 -4 1 | 0 = 4y2-4y+1=(2y-1)2, so y=½ is a zero.
Therefore we have two zeroes: 2 and ½. y=cos(3x) which cannot exceed 1, so the only solution is cos(3x)=½, 3x=π/3+2πn, 5π/3+2πn, where n is an integer, and x=π/9+⅔πn, 5π/9+⅔πn (radians).
In degrees x=20+120n, 100+120n.