sin(3π-α）=√2cos(1.5π+β）
sin[2π+(π-α)]=√2cos[π+(π/2)+β]
sin(π-α)=-√2cos[(π/2)+β]
-(-sinα)=-√2(-sinβ)
sinα=√2sinβ
√3cos(-α)=-√2cos(π+β)
√3cosα=√2cosβ
(sinα)²+(cosα)²=2（sinβ）²+（√6/3）²（cosβ)²
=2（sinβ）²+2/3[1-（sinβ）²
=4/3（sinβ）²+2/3
=1
（sinβ）²=1/4
0＜β＜π
sinβ=1/2
β=π/6或5π/6
sinα=√2sinβ=√2/2
0＜α＜π
α=π/4或3π/4
故α=π/4,β=π/6或α=3π/4,β=5π/6