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$I f 1 + \sin^{2} θ = 3 \sin θ \cos θ t h e n t h e s o l u t i o n s e t i n [0, \frac{π}{2}] i s$

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$\{\frac{π}{4}, \cos^{- 1} (\frac{1}{3})\}$

$\{\frac{π}{4}, \tan^{- 1} (\frac{1}{2})\}$

$\{\frac{π}{3}, \tan^{- 1} (\frac{1}{3})\}$

$\{\frac{π}{6}, \sin^{- 1} (\frac{1}{3})\}$

answer is B.

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Detailed Solution

$1 + \sin^{2} θ = 3 sinθ cosθ \sec^{2} θ + \tan^{2} θ = 3 tanθ 2 \tan^{2} θ - 3 tanθ + 1 = 0 (2 tanθ - 1) (tanθ - 1) = 0 tanθ = \frac{1}{2} tanθ = 1 θ = \tan^{- 1} \frac{1}{2}, \frac{π}{4}$

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