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Question

The range of θ for which the inequalitysin θ+3cos θ1 is valid if θ(π, π]is

A
θ(π3,π2]
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B
θ[π6,π2]
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C
θ(π3,π2]
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D
θ(π6,π2]
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Solution

The correct option is B θ[π6,π2]
Here, the given inequation
sin θ+3cos θ1 can solved as:
sin θ+3cos θ1Dividing by 2 in both L.H.S and R.H.S we get,12sin θ+32cos θ12sin(π6)sin θ+cos(π6)cos θ12cos( θπ6)12=cos(π3)π3 θπ6π3π6 θπ2.

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