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Question

The set of values of x satisfying the equation 1+cos3x=2cos2x can be
(where nZ)

A
nπ±π3
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B
nπ±π6
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C
nπ±π4
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D
2nπ
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Solution

The correct options are
B nπ±π6
D 2nπ
1+cos3x=2cos2x
1+4cos3x3cosx=4cos2x2
4cos3x4cos2x3cosx+3=04cos2x(cosx1)3(cosx1)=0(4cos2x3)(cosx1)=0
cos2x=34 or cosx=1
x=nπ±π6 or x=2nπ, nZ

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