find the general solution of the equation 4 cos^2x+root 3=2(root 3+1)cosx
4 cos2x + √3 =2 cosx(√3+1)
4 cos2x-2 cosx(√3+1) + √3 = 0
4 cos2x-2√3 cosx-2 cosx+√3 = 0
2 cosx(2 cosx(2 cosx-3)-1(2 cosx-√3)=0
(2 cosx-1)(2 cosx-√3)=0
Cosx=1/2 , √3/2
Cosx=√3/2=cos ∏/3 , Cosx=1/2=cos ∏/6
Therefore, x=2n∏+∏/3 or x=2n∏+∏/6
4 cos2x-2 cosx(√3+1) + √3 = 0
4 cos2x-2√3 cosx-2 cosx+√3 = 0
2 cosx(2 cosx(2 cosx-3)-1(2 cosx-√3)=0
(2 cosx-1)(2 cosx-√3)=0
Cosx=1/2 , √3/2
Cosx=√3/2=cos ∏/3 , Cosx=1/2=cos ∏/6
Therefore, x=2n∏+∏/3 or x=2n∏+∏/6