Помогите решить пожалуйста

$(tg^2\frac{\pi }{3})^{\sqrt{3x+17}}<9^2\\\\tg\frac{\pi }{3}=\sqrt3\; \; \Rightarrow \; \; tg^2\frac{\pi }{3}=(\sqrt3)^2=3\\\\3^{\sqrt{3x+17}}<(3^2)^2\\\\3^{\sqrt{3x+17}}<3^4\; \; \Rightarrow \; \; \; \sqrt{3x+17}<4\; \; \Rightarrow \; \; \left \{ {{3x+17\geq 0} \atop {3x+17<4^2}} \right. \\\\\left \{ {{x\geq -\frac{17}{3}} \atop {3x+17<16}} \right. \; \left \{ {{x\geq -5\frac{2}{3}} \atop {3x<-1}} \right. \; \left \{ {{x\geq -5\frac{2}{3}} \atop {x<-\frac{1}{3}}} \right. \; \; \Rightarrow \; \; x\in [-5\frac{2}{3}\, ;\, -\frac{1}{3})$

Cумма целых решений неравенства:

$(-5)+(-4)+(-3)+(-2)+(-1)=-15$