<s>[INST] Find the least value of $n\in N$ for which $(n-2)x^2+8x+n+4>\arcsin(\sin12)+\arccos(\cos12)$ for every $x \in \mathbb {R}$ Find the least value of $n\in N$ for which $(n-2)x^2+8x+n+4>\arcsin({\sin12})+\arccos({\cos12})$ for every $x \in \mathbb {R}$ 

$(n-2)x^2+8x+n+4>\arcsin({\sin12})+\arccos({\cos12}) $
$(n-2)x^2+8x+n+4>4\pi-12+4\pi-12$  
$(n-2)x^2+8x+n+4>8\pi-24$   
I don't know how to solve further. Please help
 [/INST] Consider $f(x)=\arcsin(\sin x)+\arccos(\cos x)$; then
$$
f'(x)=\frac{1}{\sqrt{1-\sin^2x}}\cos x+\frac{1}{\sqrt{1-\cos^2x}}\sin x=
\frac{\cos x}{\lvert\cos x\rvert}+\frac{\sin x}{\lvert\sin x\rvert}
$$
Note that $7\pi/2<12<4\pi$; one inequality is obvious; the other one is equivalent to $\pi<24/7$, but we know from Archimedes that $\pi<22/7$.
For $x\in(7\pi/2,4\pi)$ we have $\cos x>0$ and $\sin x<0$; thus $f$ is constant over $(7\pi/2,4\pi)$. Since
$$
f(4\pi-\pi/4)=0
$$
and $4\pi-\pi/4\in(7\pi/2,4\pi)$, we conclude that $f(12)=0$.
For $n\le2$, the polynomial $p(x)=(n-2)x^2+8x+n+4$ takes on negative values (it's a concave parabola for $n<1$ and a line for $n=2$).
Assuming $n>2$, you need that the discriminant of $p$ is negative:
$$
64-4(n-2)(n+4)<0
$$
that is
$$
n^2+2n-24>0
$$
so $n>4$.
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