a) 4+4cos2x=9-3cos2x b)6cos^2x+cosx-3=0
Interval 0<or equal to x <or equal to 2/pi
I assume you are solving these ....
4+4cos2x=9-3cos2x
7cos 2x = 5
cos 2x = 5/7
2x = .77519... or 2x = 2π - .77519 = 5.50799
x = .3876 or x = 2.754
but according to your interval, we only need x = .3876
b) 6cos^2x+cosx-3=0
let cosx = u, then we have
6u^2 + u - 3 = 0
u = (-1 ± √73)/12 = .62866... or u = a negative, which lies outside our domain, so
cosx = .62866...
x = appr .891