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

Question

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

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$(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

Answer 1

Since $$\arcsin\sin12+\arccos\cos12=0$$ and $n=2$ is not valid, we need $$4^2-(n-2)(n+4)<0.$$ Can you end it now?

I got $n>4$ or $n<-6.$

Answer 2

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 $n>4$.