Please help me factor this?

Question

2 cos(x)^2 - 11 cos(x) + 5 = 0

-- Here are my steps --

(2 cos(x) - 10) (2 cos(x) - 1) = 0
2 ( cos(x) - 5) = 0 2 cos(x) - 1 = 0
2 = 0 cos(x) - 5 = 0 2 cos(x) - 1 = 0
cos(x) = 5 cos(x) = 1

What do I do now? None of this stuff matches the answers in my multiple choice packet. Gosh, I hate trig :(

Answer 1

put cosx = z an find the roots of

2z^2-11z+5=0 z=((11+-sqrt(121-40))/4 = (11+-9)4

z=5 and z =1/2 so cos x=5 (CAN´T BE) and cos x= 1/2 which gives x= pi/3 and x= 5pi/3 in the interval (0,2pi)
The factoring would be 2*(cosx-5)*(cos x-1/2)=(cos x-5)*(2cosx-1)

Answer 2

2 cos(x)^2 - 11 cos(x) + 5 = 0

2 cos(x)^2 - 10 cos(x) - cos(x) + 5 = 0
2 cos(x) { cos(x) -5} -{ cos(x) -5} = 0
Now { cos(x) -5} is common in the two terms.
Thus
{ cos(x) -5}*{2 cos(x) -1} =0