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已知0由0<a-b<π/2 即0<2a-2b<π(1)π/2<a+2b<3π/2(2)(1)+(2)得:π/2<3a<5π/2∴π/6<a<5π/6.(2)-(1)得:π/2<3b<π,∴π/6<b<π/3即π/3<a+b<7π/6.