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?

The smallest positive integer is $1$. That doesn't satisfy the condition, because
$$ \sin^2 x + \cos x + 1 > 1 + \cos x $$
has equality when $\sin^2x = 0$, which happens, among other places, at $x=0$.
The next positive integer is $2$. Does that work?
$$ \sin^2 x + 2\cos x + 4 > 1 + \cos x $$
holds if and only if
$$ \sin^2 x + \cos x + 3 > 0 $$
which is easily true -- since $\sin^2 x$ is never less than $0$ and $\cos x$ is never less than $-1$, the left-hand side is always $\ge 2$.
So the answer is $$ \Huge 2 $$
