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Find all angles $\theta,$ $0 \le \theta \le 2 \pi,$ with the following property: For all real numbers $x,$ $0 \le x \le 1,$

$$x^2 \cos \theta - x(1 - x) + (1 - x)^2 \sin \theta > 0.$$

I am not exactly sure how to solve this problem. A friend gave me the suggestion to use the substitution $\sin^2(y)=x,\cos^2(y)=1-x$. Then, $\sin^4(y)\cos(\theta)-\sin^2(y)\cos^2(y)+\cos^4(y)\sin(\theta) > 0$. I tried to exploit symmetry by dividing by $\sin^2(y)\cos^2(y)$ to get $\tan^2(y)\cos(\theta)+\cot^2(y)\sin(\theta)>1$. I am not sure how to continue from here. Anything I tried from here was fruitless. What should I do to solve this problem?

Albus Dumbledore
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djdumpling
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1 Answers1

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Hint

$$f(x)=x^2 \cos \theta - x(1 - x) + (1 - x)^2 \sin \theta > 0\Leftrightarrow$$ $$f(x)=x^2(\sin \theta+\cos \theta+1)-x(1+2\sin \theta)+\sin \theta>0.$$

Let's look at the discriminant, $$\Delta=(1+2\sin \theta)^2-4(\sin \theta+\cos \theta+1)\sin \theta$$ $$\Delta=1-2\sin (2\theta)\to -1\le \Delta\le 3.$$

First case of solutions,

  • $\Delta<0$ and $\sin \theta+\cos \theta+1 >0.$

Second case,

  • $\Delta\ge 0$: $x_1\le x_2$ are the roots of $f(x)$,

a) If $\sin \theta+\cos \theta+1 >0$ then $x_2\le0$ or $x_1\ge 1$;

b) If $\sin \theta+\cos \theta+1 <0$ then $[0,1]\subset[x_1,x_2]$.

Arnaldo
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