The values of θ∈(0,2π) for which 2sin2θ−5sinθ+2>0, are

2sin2θ−5sinθ+2>0 ⇒(sinθ−2)(2sinθ−1)>0⇒sinθ 0(no solutions)θ∈(0,π6)∪(5π6,2π)

The values of θ∈(0,2π) for which 2sin2θ−5sinθ+2>0, are

2sin2θ−5sinθ+2>0 ⇒(sinθ−2)(2sinθ−1)>0⇒sinθ 0(no solutions)θ∈(0,π6)∪(5π6,2π)

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The values of $θ \in (0, 2 π)$ for which $2 \sin^{2} θ - 5 \sin θ + 2 > 0,$ are

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$(0, \frac{π}{6}) \cup (\frac{5 π}{6}, 2 π)$

$(\frac{π}{8}, \frac{5 π}{6})$

$(0, \frac{π}{8}) \cup (\frac{π}{6}, \frac{5 π}{6})$

$(\frac{41 π}{48}, π)$

answer is A.

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Detailed Solution

$2 \sin^{2} θ - 5 \sin θ + 2 > 0$

$\Rightarrow (\sin θ - 2) (2 \sin θ - 1) > 0$

$\Rightarrow \sin θ < \frac{1}{2}$ $\sin θ - 2 > 0 (n o s o l u t i o n s)$

$θ \in (0, \frac{π}{6}) \cup (\frac{5 π}{6}, 2 π)$

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The values of θ∈(0,2π) for which 2sin2θ−5sinθ+2>0, are

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The values of $θ \in (0, 2 π)$ for which $2 \sin^{2} θ - 5 \sin θ + 2 > 0,$ are