Why are my calculations off?

Question

solve 5cos(2x)=5cos^2(x)-4      0 ≤  x <  2π

Answer 1

5cos(2x)=5cos^2(x)-4 

Double angle formula : cos2A =2cos²A-1

5(2cos²x-1)=5cos²x-4 

10cos²x-5 =5cos²x-4

5cos²x=1

cos²x=1/5

cosx =±1/√5

cosx =arccos(1/√5)  and  cosx =arccos(-1/√5) 

x = 2nπ±arccos(1/√5) and x = 2nπ±arccos(-1/√5)

x = 2nπ±63.435°   and x = 2nπ±116.565°

x = 2nπ±0.3524π   and x = 2nπ±0.6476π

At n=0 : x = 2(0)π - 0.3524π =-1.10654 and x = 2(0)π - 0.6476π =-2.033464

At n=1 : x = 2(1)π - 0.3524π =5.173464 and x = 2(1)π - 0.6476π =4.246536