Find a rational range and solve by adding a cosine sketch

$y(x+1)=x^2+2x+3$ gives $x^2+(2-y)x+(3-y)=0$. No roots requires $(2-y)^2-4(3-y)<0$. Thus $4-4y+y^2-12+4y<0$, or $y^2-8<0$, hence $(y-2\sqrt2)(y+2\sqrt2)<0$. Therefore the missing interval is $-2\sqrt2<y<2\sqrt2$.

$y=x+1+2/(x+1)$, so asymptotes are x=−1,y=x+1. The only axial intercept is (0,3), since the numerator (x+1)²+2 has no real zero. The turns are $(-1-\sqrt2,-2\sqrt2)$, maximum≈(−2.41,−2.83), and $(-1+\sqrt2,2\sqrt2)$, minimum≈(0.414,2.83). All source rounded coordinates are retained.

$3(x+1)\cos x-1=x^2+2x+2$ gives $3\cos x=(x^2+2x+3)/(x+1)$. x=−1 is not a solution of the original equation, so division does not lose a root.

Add y=3 cosx on−π≤x≤π. The source concludes x=0 or x≈0.335 to 3 significant figures. Both are retained. The left branches approach closely near x=−3 but do not cross; the supplementary zoom keeps that near approach from being mistaken for an additional solution.