The set of values of x satisfying the equation 1+cos3x=2cos2x can be (where n∈Z)
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 Bnπ±π6 D2nπ 1+cos3x=2cos2x ⇒1+4cos3x−3cosx=4cos2x−2 ⇒4cos3x−4cos2x−3cosx+3=0⇒4cos2x(cosx−1)−3(cosx−1)=0⇒(4cos2x−3)(cosx−1)=0 ⇒cos2x=34 or cosx=1 x=nπ±π6 or x=2nπ,n∈Z