f(x)=2(sinx*cosπ/3+cosx*sinπ/3)-2sinx
=2(sinx*1/2+cosx*√3/2)-2sinx
=sinx+√3cosx-2sinx
=√3cosx-sinx
=2(√3/2cosx-1/2sinx)
=2sin(π/3-x)
因为x属于[-π/2,0],所以 π/3-x属于[π/3,5π/6]
所以f(x)属于[1/2,1]
因为cosx=√3/3,x属于[-π/2,0]
所以sinx=-√6/3
带回得f(x)=(3+√6)/3
f(x)=2(sinx*cosπ/3+cosx*sinπ/3)-2sinx
=2(sinx*1/2+cosx*√3/2)-2sinx
=sinx+√3cosx-2sinx
=√3cosx-sinx
=2(√3/2cosx-1/2sinx)
=2sin(π/3-x)
因为x属于[-π/2,0],所以 π/3-x属于[π/3,5π/6]
所以f(x)属于[1/2,1]
因为cosx=√3/3,x属于[-π/2,0]
所以sinx=-√6/3
带回得f(x)=(3+√6)/3