Different Solutions

hi frnds, I am confused. i am getting two different answers .help me out.
Solve;
4sin(2x) + 3 cos (2x) =5
My solution is:
4(2sinx cos x) + 3 (2cos²x - 1) =5
→8sinx cosx + 6cos²x = 8
Divide both sides by cos²x
→8tanx + 6 = 8 (1+ tan²x)
→8tan²x - 8tanx + 2 =0
→(2tanx- 1)² =0
→X= n (pi) + tan^-1(1/2) =0
But the answer given is n(pi)+ (1/2) tan^-1(4/3)

in the book , first they divide both sides by 5 then they form an eqn of form Cos (a-2x), where a = cos^-1(3/5) where they get X = n (pi) + (1/2) cos^-1(3/5) which is same as n(pi) + (1/2) tan^-1 (4/3) the method is fine but i am getting different answers.