sin(a+π/3)+sina=-4√3/5,-π/2

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∵－π/2＜a＜0,∴π/6－π/2＜a＋π/6＜π/6,∴－π/3＜a＋π/6＜π/6.
∵sin（a＋π/3）＋sina＝－4√3/5,∴2sin（a＋π/6）cos（π/6）＝－4√3/5,
∴sin（a＋π/6）＝－4/5,结合－π/3＜a＋π/6＜π/6,得：
cos（a＋π/6）＝√（1－16/25）＝3/5.
∴cosa＝cos［（a＋π/6）－π/6］＝cos（a＋π/6）cos（π/6）＋sin（a＋π/6）sin（π/6）
＝（3/5）×（√3/2）＋（－4/5）×（1/2）＝3√3/10－4/10＝（3√3－4）/10.
======以下答案可供参考======
供参考答案1:
√3cosa / 2 + 3sina/ 2 = -4√3/5
√3sin(a+β）= -4√3/5 ，tanβ = √3/3
sin（a + π/6） = -4/5
√3sina /2 + cosa /2 = -4/5
100cos²a + 80cosa -11= 0
cosa =