cos3θ-cosθ=0 θ∈[0,π]

cos3θ-cosθ=0 θ∈[0,π]
cos3θ-cosθ=0 θ∈[0,π]

cos3θ-cosθ=0 θ∈[0,π]
cos3θ-cosθ=0 θ∈[0,π]
cos3θ-cosθ=cos(2θ+θ)-cosθ=cos2θcosθ-sin2θsinθ-cosθ
=(cos2θ-1)cosθ-sin2θsinθ=(1-2sin^2θ-1)cosθ-2sinθcosθsinθ
=-2sin^2θcosθ-2sin^2θcosθ=-4sin^2θcosθ=0
所以sinθ=0或cosθ=0,
θ∈[0,π],所以θ=0°,或π,或π/2

三倍角公式：cos3x=4cos³x-3cosx
该题：
4cos³θ-4cosθ=0
cosθ(cos²θ-1)=0
cosθ=0或cosθ=±1
∵θ∈[0,π]
∴θ=0或θ=π或θ=π/2