If 0≤x≤2π and |cos x|≤sinx, then

If 0x2π and |cos x|sinx, then

  1. A

    x0,π4

  2. B

    xπ4,2π

  3. C

    π4,3π4

  4. D

    [0,π]

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    Solution:

    We have, |cosx|sinxsinx0(|cosx|0)

    x(π,2π) if x=2π,|cos2π|sin2π

    Which is not possible x0,π2,then

    |cosx|sinxxπ4,π2

    If xπ2,π, then |cosx|sinx

    cosxsinx  (cosx<0)

    x(π2,3π4]

    x[π4,π2](π2,3π4]

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