Given,

3 cos 4 - 5 cos 2 -2 sin 6 + 2 = 0

cos 2 [3 cos 2 - 5] -2 sin 6 + 2 = 0

(1 - sin 2 ) (3 - 3sin 2 - 5)-2sin 6 + 2 = 0

(sin 2 - 1) (3 sin 2 + 2) -2 sin 6 + 2 = 0

Let sin 2 = t

(t-1) (3t + 2) – 2t 3 + 2 = 0

(t-1) (3t + 2) - 2(t 3 - l) = 0

We know that a 3 - b 3 = (a - b) (a 2 + ab + b 2 )

(t-l) (3t + 2) -2(t-1) (t 2 +t + l) = 0

(t- l) [3t+ 2 - 2(t 2 + t + 1)1= 0

(t-1) [2t 2 -t] =0

For sin = 0,

Total solution = 9

Hence this is the required option.

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