高中数学,已知三角函数求角sin(x-π/6)= -√3/2 x属于(0,2π)求x集合拜托各位了,原题上有讲解没看懂

首先,你要对-√3/2这个数字有感觉.知道sin(-π/3)=-√3/2 所以x-π/6 + 2nπ=-π/3 或者π-( x-π/6) +2nπ= -π/3 因为sin(t)=sin(π-t) x-π/6 + 2nπ=-π/3 :x + 2nπ= -π/6 因为x属于(0,2π),所以n=-1 x=-π/6 + 2π =11π/6 π-( x-π/6) +2nπ= -π/3:7π/6-x +2nπ= -π/3 7π/6 + π/3 + 2n π = x 9π/6 + 2n π= x 3π/2 +2nπ=x 因为x属于(0,2π),所以n=0 x= 3π/2