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Question

Solve:|cosx|=cosx2sinx

A
x=(2n+1)π+π4
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B
x=(2n+1)π+π2
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C
x=2nπ+π4
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D
x=2nπ+π2
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Solution

The correct option is A x=(2n+1)π+π4
We have, |cosx|=cosx2sinx
Case 1)
if cosx0
cosx=cosx2sinx

sinx=0x=nπ

x=2nπ
Case 2)
If cosx<0
cosx=cosx2sinx

cosxsinx=1

tanx=1x=nπ+π4

x=(2n+1)π+π4

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