Interesting Trig Equation

$cos(2x)=1-2sin^2x$ , so the equation becomes

$1-2\sin^2x+2\sin x-\sqrt{3}\sin x=1-\sqrt{3}$

$2\sin^2x+(\sqrt{3}-2)\sin x-\sqrt(3)=0$

$(2\sin x + \sqrt{3})(\sin x -1 )=0$

Therefore, $\sin x = -\frac{\sqrt{3}}{2}$ or $\sin x = 1$ .

On the interval $[0^{\circ}, 270^{\circ}]$ , the only solutions are $240$ and $90$ , so the answer is $240+90=\fbox{330}$ .