Minimum value of this trigonometric function?! PLEASE HELP!

cos^2x - 6sinx cosx + 3sin^2x + 2

(a) 4 + root10
(b) 4 - root10
(c) 0
(d) none

cos^2(x) - 6sin(x)cos(x) + 3sin^2(x) + 2

= cos^2(x) + sin^2(x) + 2sin^2(x) + 2 - 6sin(x) cos x

= 2sin^2(x) + 3 - 3sin(2x)

= 2sin^2(x) - 1 + 4 - 3sin(2x)

= - cos(2x) - 3sin(2x) + 4

= 4 - [cos(2x) + 3sin(2x) ]

make into single trignometric function

= 4 - [a sin A cos(2x) + b cos A sin(2x) ]

where a sin A = 1 and a cos A = 3

a^2 sin^2(A) + a^2 cos^2(A) = 1 + 9 = 10

a^2 = 10

a = √10

=> 4 - √10[sin A cos(2x) + cos A sin(2x) ]

= 4 - √10[sin(A + 2x) ]

since minimum value of sin(A + 2x) = -1,

The minimum value of given function = 4 - √10(-1)

= 4 + √10