The value of θ(0 ≤ θ ≤ 90°) satisfying 2sin2θ = 3cosθ is?

The value of θ(0 ≤ θ ≤ 90°) satisfying 2sin2θ = 3cosθ is? Correct Answer 60°

$$\eqalign{ & {\text{Hit and Trial method}} \cr & {\text{Put }}\theta = {60^ \circ }{\text{option A}} \cr & \Rightarrow 2{\sin ^2}{60^ \circ } = 3\cos {60^ \circ } \cr & \Rightarrow 2{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} = 3\left( {\frac{1}{2}} \right) \cr & \Rightarrow \frac{3}{2} = \frac{3}{2}{\text{ }}\left( {{\text{LHS}} = {\text{RHS}}} \right) \cr} $$

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