Using the substitutions $$\tan\frac x2=t,\quad x=2\arctan t,\quad \cos x=\frac{1-t^2}{1+t^2},\quad \sin x=\frac{2t}{1+t^2}, $$ rewrite the inequality in question as $$d(t):=\frac{2 t \left(39 t^6-29 t^4-95 t^2-75\right)}{3 \left(t^2+1\right)^2 \left(13 t^4-10 t^2+25\right)}+2 \tan ^{-1}t\ge0$$ for $t\in[0,1]$.
It remains to note that $d(0)=0$ and $$d'(t)=\frac{128 t^6 \left(13 t^4+40 t^2+75\right)}{\left(t^2+1\right)^3 \left(13 t^4-10 t^2+25\right)^2}\ge0$$ for all real $t$.