Answer :

Given,


2sin2θ = 3cos θ


We know that – sin2θ = 1 – cos2θ



2(1 – cos2θ) = 3cos θ


2 – 2cos2θ = 3cos θ


2cos2θ + 3cos θ - 2 = 0


2cos2θ + 4cos θ - cos θ - 2 = 0


2cos θ (cos θ+ 2) +1 (cos θ + 2) = 0


(2cos θ + 1)(cos θ + 2) = 0


cos θ [-1,1] , for any value θ.


So, cos θ ≠ - 2


2 cos θ - 1 = 0


cos θ = 1/2




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