若sin(x+π/6)+cos(x+π/6)=(根号2)/2,其中x∈(-π/6,7π/12),则sin2x=

问题描述:

若sin(x+π/6)+cos(x+π/6)=(根号2)/2,其中x∈(-π/6,7π/12),则sin2x=

sin(x+π/6)+cos(x+π/6)=(√2)/2
两边平方:1+2sin(x+π/6)cos(x+π/6)=1/2
∴1+sin﹙2x+π/3﹚=1/2
∴sin﹙2x+π/3﹚=-1/2,说明2x+π/3为第三象限或者第四象限的角……①
∵-π/6<x<7π/12
∴-π/3<2x<7π/6
∴0<2x+π/3<3π/2……②
由①②知,2x+π/3为第三象限的角
∴由sin﹙2x+π/3﹚=-1/2知,2x+π/3=7π/6
∴2x=5π/6
∴sin2x=sin﹙5π/6﹚=1/2