If |2sinθ−cosec θ|≥1, then complete set of values of θ is
(where n∈Z)
Open in App
Solution
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≥12orcos2θ≤−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}