1)
let 20 = θ
we are asked to find cosθ(4sinθ-√3)sinθ
we have 3sinθ-4sin3θ = √32 ....(1)
and 4cos3θ - 3cosθ = 12 .....(2)
from (2), 4sinθ - √3 = 2sinθ(4sin2θ-1)
=> cosθ(4sinθ-√3)sinθ = 2cosθ(4sin2θ-1)1 = 2(3cosθ-4cos3θ)1
= -1