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Question

If log0.5sinx=1log0.5cosx then number of values of x[2π,2π] is

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 D 4
log0.5cosxsinx=1
cosxsinx=12
2sinxcosx=1
sin2x=1
2x=(2k+1)Π2
No.ofvaluesofxsatisfying=4

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