Đáp án:

`Sin2x -3\sqrt{3}.(Sinx+Cosx)+5=0`

`<=> 2Sinx.Cosx -3\sqrt{3}.(Sinx+Cosx) +5 = 0 (1)`

Ta có : `Sin^2x +Cos^2x = 1`

`-> (Sinx+Cosx)^2 -2Sinx.Cosx = 1`

`-> 2SinxCosx = (Sinx+Cosx)^2 -1`

`(1) <=> (Sinx+Cosx)^2  -3\sqrt{3}.(Sinx+Cosx) +4 = 0`

`-> Sinx+Cosx = (\sqrt{11}+3\sqrt{3})/2` (vô lí ) hoặc `Sinx+Cosx = (-\sqrt{11}+3\sqrt{3})/2`

`-> \sqrt{2}.Sin(x+π/4) = (3\sqrt{3}-\sqrt{11})/2 `

`-> Sin(x+π/4) = (\sqrt{22}+3\sqrt{6})/4`

`-> x+π/4 = arcsin((\sqrt{22}+3\sqrt{6})/4) +k2π` hoặc `x+π/4 = π-arcsin((\sqrt{22}+3\sqrt{6})/4) +k2π`

`-> x = arcsin((\sqrt{22}+3\sqrt{6})/4) +k2π -π/4` hoặc `x = (3π)/4 -arcsin((\sqrt{22}+3\sqrt{6})/4) +k2π (k in ZZ )`