Đáp án:
`\sqrt{3}.tan(2x-36^0)-1=0(90^{0}<x<180^{0})`
`<=>\sqrt{3}.tan(2x-36^0)=1`
`<=>tan(2x-36^0)=(1)/(\sqrt{3})`
`<=>tan(2x-36^0)=tan30^0`
`<=>2x-36^0=30^0+k.180^0;k∈ZZ`
`<=>2x=66^0+k.180^0;k∈ZZ`
`<=>x=33^0+k.90^0;k∈ZZ`
Ta có: `90^0<x<180^0`
`<=>90^0<33^0+k.90^0<180^0`
`<=>57^0<k.90^0<147^0`
`<=>57/90 =19/30≈0,63<k<147/90=49/30≈1,63`
`k∈ZZ->k=1`
`->x=33^0+90^0=123^0`
Vậy `S={123^0}`
_____________________________
`tanx=tan\alpha`
`->x=\alpha+k\pi` hay `x=\alpha+k.180^0(k∈ZZ)`
