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Question

Solve cosθ+cos3θ2cos2θ=0

A
θ=(2n+1)π4,nϵZ
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B
θ=mπ,mϵZ
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C
θ=2mπ,mϵZ
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D
θ=(2n+1)π8,nϵZ
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Solution

The correct options are
A θ=(2n+1)π4,nϵZ
B θ=2mπ,mϵZ
We have cosθ+cos3θ2cos2θ=0
2cos2θcosθ2cos2θ=0
2cos2θ(cosθ1)=0
cos2θ=0 or cosθ1=0
2θ=(2n+1)π2,nZ or θ=2mπ,mZ
θ=(2n+1)π4,nZ or θ=2mπ,mZ

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