已知sin(x+π/3)+sin x=(-4√3)/5,(-π/2)<x<0,求cos x的值

解:sin(x+π/3)+sinx=(-4√3)/5
则:
(sinxcosπ/3+sinπ/3cosx)+sinx=(-4√3)/5
(3/2)sinx+(√3/2)cosx=(-4√3)/5
√3sinx+cosx=-8/5
则:cosx=-√3sinx-8/5 -----(1)

又sin^2(x)+cos^2(x)=1
则将(1)代入,得:
sin^2(x)+[√3sinx+8/5]^2=1
4sin^2(x)+16√3/5sinx+39/25=0
由于(-π/2)<x<0
则: sinx<0
则: sinx=
{-16√3/5-√[(-16√3/5)^2-4*4*39/25]}/2*4
=-(15+4√3)/10
则:
cosx=-√3sinx-8/5
=(15√3-4)/10

sin(x+π/3)+sinx
=sinxcosπ/3+cosxsinπ/3+sinx
=1/2sinx+√3/2cosx+sinx
=3/2sinx+√3/2cosx=(-4√3)/5
∴√3sinx+cosx=-8/5
∴sinx=(-8/5 -cosx)/√3
∵(sinx)^2+(cosx)^2=1
∴[(-8/5 -cosx)/√3]^2+(cosx)^2=1
整理后：100(cosx)^2+80cosx-11=0
∴cosx=(-4+3√3)/10 或cosx=(-4-3√3)/10
∵(-π/2)<x<0
∴cosx=(-4+3√3)/10