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Solve log(|sin x|) (1+cos x)=2....

Solve `log_(|sin x|) (1+cos x)=2`.

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The correct Answer is:
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`log_(|sin x|) (1+cos x)=2`
`rArr 1+cos x=sin^(2) x`
`rArr 1+cos x=(1-cos x) (1+ cos x)`
`rArr 1+cos x =0` or `1-cos x=1`
`rArr cos x=-1` or `cos x=0`
`rArr sin x=0` or `|sin x|=1`
Both of these are not possible as base of logarithm cannot be either 0 or 1.
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