<s>[INST] Trigonometric Manipulation with inequality:$x^2 \cos \theta - x(1 - x) + (1 - x)^2 \sin \theta > 0.$ 
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?
 [/INST] 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 $[0,1]\subset[x_1,x_2]$.
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