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Question

If cosx=1sin2x,0xπ, then possible value of x is

A
π
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B
0
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C
tan12
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D
3π
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Solution

The correct options are
A tan12
B 0
cosx=1sin(2x) x[0,x]
=12sinxcosx {1=sin2x+cos2xxR}
=sin2x2sinxcosx+cos2x
=[sinxcosx]2 {x=|x|}
cosx=|sinxcosx|
Case I : 1sinxcosx:x[0,x]
|sinxcosx|=sinxcosx
cosx=sinxcosx
2cosx=sinxtanx=2
x=tan1(2) [x[0,x]]
Case III:
sinx<cosx;x[0,x]
|sinxcosx|=cosxsinx
cosx=cosxsinx
sinx=0
x=0,π
But x=π is rejected as cos(x)=1 and 01
only x=0
Finally : x=tan1(2),0

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