已知cos(π/2+α)=根号2cos(3π/2-β),根号3sin(3π/2+α)=-根号2sin(π/2-β),且0

已知cos(π/2+α)=根号2cos(3π/2-β),根号3sin(3π/2+α)=-根号2sin(π/2-β),且0

cos(π/2 + α) = √2cos(3π/2 - β) => -sinα = -√2sinβ => sinα = √2sinβ ①,√3sin(3π/2 + α) = -√2sin(π/2 - β) => -√3cosα = -√2cosβ => √3cosα = √2cosβ ②,把① + ②,可得sin2α + 3cos2α = 2sin2β + 2cos2β = 2 => 1 + 2cos2α = 2 => 2cos2α = 1 => cos2α = 1/2 => cosα =±√2/2 => α = π/4 + kπ/2,k∈Z,因为0 < α < π,所以只能是k = 0或者1；
1）当k = 0时,α = π/4,所以sinα = cosα = √2/2 => sinβ = 1/2和cosβ = √3/2 => β = π/6 + 2kπ,k∈Z,因为0 < β < π,所以只能是k = 0,β = π/6；
2）当k = 1时,α = 3π/4,所以sinα = √2/2和cosα = -√2/2 => sinβ = 1/2和cosβ = -√3/2 => β = 5π/6 + 2kπ,k∈Z,因为0 < β < π,所以只能是k = 0,β = 5π/6；
综上所述,α = π/4,β = π/6；或者α = 3π/4,β = 5π/6.