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Question

Number of values of θϵ[0,2π] satisfying the equation cotxcosx=1cotx.cosx

A
1
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B
2
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C
3
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D
4
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Solution

The correct option is C 3
cosxsinxcosx=1cos2xsinx

cosxcosxsinx=sinxcos2x

cos2x+cosxcosxsinxsinx=0

cosx(1+cosx)sinx(1+cosx)=0

(cosxsinx)(1+cosx)=0

x=π4,5π4,π

Hence, 3 solutions exist.

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