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Q.

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

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a

(0,π6)∪(5π6,2π)

b

(π8,5π6)

c

(0,π8)∪(π6,5π6)

d

(41π48,π)

answer is A.

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Detailed Solution

2sin2θ−5sinθ+2>0 ⇒(sinθ−2)(2sinθ−1)>0⇒sinθ<12      sinθ−2>0(no solutions)θ∈(0,π6)∪(5π6,2π)
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The values of θ∈(0,2π)  for which 2sin2θ−5sinθ+2>0,  are