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B
⋃n∈Z[nπ−π3,nπ+π3]∪{(2n+1)π2}−{nπ}
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C
⋃n∈Z(nπ−π3,nπ+π3)
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D
⋃n∈Z(nπ−π6,nπ+π6)
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Solution
The correct option is A⋃n∈Z[nπ−π6,nπ+π6]∪{(2n+1)π2}−{nπ} cosec θ is not defined for nπ,∀n∈Z |2sinθ−cosec θ|≥1 ⇒|2sin2θ−1|≥|sinθ| ⇒2cos22θ≥1−cos2θ ⇒2cos22θ+cos2θ−1≥0 ⇒(2cos2θ−1)(cos2θ+1)≥0 ⇒cos2θ≥12orcos2θ≤−1
Now, using the graph 2θ∈[−π3,π3] and period of cos2θ is π.
If cos2θ≤−1, then θ=(2n+1)π2
Hence, θ∈⋃n∈Z[nπ−π6,nπ+π6]∪{(2n+1)π2}−{nπ}