6 sin 2θ tan θ = 4, where 0º < θ < 90º
sin 2θ = 2sin θ cos θ and tanθ = \(\frac{sinθ}{cosθ}\)
= 6 x 2sin θ cos θ x \(\frac{sin θ}{cos θ} = 4\)
= \(sin^2 θ = 4\)
= \(sin^2 θ = \frac{4}{12}=\frac{1}{3}\)
=\(sin θ = \frac{\sqrt1}{3}=\frac{1}{\sqrt3}\)
= \(θ = sin^{-1}(\frac{1}{\sqrt3})\)
∴ θ = 35.26º
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