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Question

The number of values of x[0,nπ],nI that satisfy log|sinx|(1+cosx)=2 is


A

0

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B

n

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C

2n

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D

None of these

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Solution

The correct option is A

0


The equation is meaningful if |sinx|0 or1 and 1+cosx0

so xkπ,k=0,1,n,and x(2k+1)π2,k=0,...n1.

log|sinx|(1+cosx)=2
1+cosx=|sinx|2=sin2x=1cos2x(1+cosx)(cosx)=0cosx=0 or cosx=1cosx=0x=(2k+1)π2.
cosx=1x=kπ.
So there is no x which satisfy the given equation.


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