Smallest positive integral value of $a$ such that ${\sin}^2 x+a\cos x+{a}^2>1+\cos x$ holds for all real $x$ 
If the inequality $${\sin}^2 x+a\cos x+{a}^2>1+\cos x$$ holds for all $x \in \Bbb R$ then what's the smallest positive integral value of $a$?

Here's my approach to the problem $$\cos^2 x+(1-a)\cos x-a^2<0$$
Let us consider this as a quadratic form respect to $a$.
Applying the quadratic formula $a=\frac{-\cos x\pm\sqrt{5\cos^2 x+4\cos x}}2 $
and substituting $\cos x$ with $1$ and $-1$
we get 3 values of where the graph should touch the x axis $-2,0,1$
How should I proceed now?

$$ 
\cos^2{x}+(1-a)\cos{x}-a^2\,\lt0 \quad\&\quad \left|\,\cos{x}\,\right|\,\le1 \\[6mm] 
-1\le\,\cos{x}=\frac12\left(\,-(1-a)\pm\sqrt{(1-a)^2+4a^2}\,\right)\,\le+1 \\ 
-1-a\,\le\,\pm\sqrt{5a^2-2a+1}\,\le\,+3-a \\[6mm] 
\begin{align} 
&\text{For}\,\colon\,\,\,\pm\sqrt{5a^2-2a+1}\,\ge\,-1-a \implies 5a^2-2a+1=a^2+2a+1 \\ 
&\implies 4a^2-4a=a^2-a=0\implies a=0,\,a=1\quad\color{red}{a\in[0,1]} \\[6mm] 
&\text{For}\,\colon\,\,\,\pm\sqrt{5a^2-2a+1}\,\le\,+3-a \implies 5a^2-2a+1=a^2-6a+9 \\ 
&\implies 4a^2+4a-8=a^2+a-2=0\implies a=-2,\,a=1\quad\color{blue}{a\in[-2,1]} \\[6mm] 
\end{align} 
$$ 
Hence, for the inequality $\,\{\cos^2{x}+(1-a)\cos{x}-a^2\,\lt0\}\,$ to hold for all $\,x\in\mathbb{R}\,$ and the operator without equal (less than only), Then: 
$$ \boxed{ \quad \color{blue}{a \in (-\infty,\,-2)\,\wedge\,(1,\infty) } \quad } $$ 
And for $\,a\in\mathbb{N}^{+}\implies\color{red}{a=2}\,$

 

