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Question

The smallest positive value of x which satisfies the equation logcosxsinx+logsinxcosx=2 is

A
π2
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B
π3
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C
π4
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D
π6
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Solution

The correct option is C π4
logcosxsinx+logsinxcosx=2
Let logcosxsinx=t
t+1t=2t22t+1=0
(t1)2=0t=1
logcosxsinx=1 or sinx=cosxtanx=1
x=π4

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