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Question

# If |2sinθ−cosec θ|≥1, then complete set of values of θ is (where n∈Z)

A
[nππ6,nπ+π6]{nπ}
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B
[nππ6,nπ+π6]{(2n+1)π2}
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C
[nππ6,nπ+π6]
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D
[nππ6,nπ+π6]{(2n+1)π2}{nπ}
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Solution

## The correct option is D [nπ−π6,nπ+π6]∪{(2n+1)π2}−{nπ}Given : |2sinθ−cosec θ|≥1 For cosec θ to be defined, we get θ≠nπ,∀ n∈Z Now, |2sinθ−cosec θ|≥1⇒|2sin2θ−1|≥|sinθ|⇒|−cos2θ|≥|sinθ| Squaring on both sides, we get ⇒cos22θ≥sin2θ⇒2cos22θ≥1−cos2θ⇒2cos22θ+cos2θ−1≥0⇒(2cos2θ−1)(cos2θ+1)≥0⇒cos2θ≥12 or cos2θ≤−1 As cos2θ∈[−1,1], so cos2θ≥12,cos2θ=−1 For cos2θ≥12, using the graph of cosx, we get 2θ∈[−π3,π3]⇒2θ∈[2nπ−π3,2nπ+π3]⇒θ∈[nπ−π6,nπ+π6] Also, cos2θ=−1⇒2θ=(2n+1)π⇒θ={(2n+1)π2} Hence, θ∈[nπ−π6,nπ+π6]∪{(2n+1)π2}−{nπ}, n∈Z

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