Question

已知sin(a+π/3)+sina=-4√3/5,-π/2<a<0,求cosa的值

请告诉我答案及解题过程！！！谢谢！！！

Answer 1

sin(a+π/3)+sina=-4√3/5
sinacosπ/3+cosasinπ/3+sina=-4√3/5
(3/2)sina+(√3/2)cosa=-4√3/5
sina=(2/3)*(-4√3/5-√3/2*cosa)=-(8√3/15+√3/3*cosa)
代入(sina)^2+(cosa)^2=1
令x=cosa
[-(8√3/15+√3/3*x)]^2+x^2=1
64/75+16x/15+x^2/3+x^2=1
(4/3)x^2+(16/15)x-11/75=0
x=(-4±3√3)/10
-π/2<a<0
cosa>0
所以cosa=(-4+3√3)/10