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Question

If |2sinθcosec θ|1, then complete set of values of θ is
(where nZ)

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Solution

Given : |2sinθcosec θ|1
For cosec θ to be defined, we get
θnπ, nZ
Now,
|2sinθcosec θ|1|2sin2θ1||sinθ||cos2θ||sinθ|
Squaring on both sides, we get
cos22θsin2θ2cos22θ1cos2θ2cos22θ+cos2θ10(2cos2θ1)(cos2θ+1)0cos2θ12 or cos2θ1
As cos2θ[1,1], so
cos2θ12,cos2θ=1


For cos2θ12, using the graph of cosx, we get
2θ[π3,π3]2θ[2nππ3,2nπ+π3]θ[nππ6,nπ+π6]
Also,
cos2θ=12θ=(2n+1)πθ={(2n+1)π2}

Hence, θ[nππ6,nπ+π6]{(2n+1)π2}{nπ}, nZ

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