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Question

cos6θ+cos4θ+cos2θ+1=0;0θπ.

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Solution

(cos6θ+cos4θ)+(1+cos2θ)=0,0θπ
2cos5θcosθ+2cos2θ=0
or 2cosθ(cos5θ+cosθ)=0
or 4cosθcos2θcos3θ=0
cosθ=0 θ=(n+12)π
n=0, θ=π/2
cos2θ=0 2θ=(n+12)π
θ=(2n+1)π/4
n=0,θ=π,n=1,θ=3π/4,
cos3θ=0
3θ=nπ+π/2
θ=(2n+1)π/6
n=0,θ=π/6,n=1,θ=π/2,
n=2,θ=5π/6.
θ=30o,45o,90o,135o,150o.

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