Smallest positive integral value of $a$ such that ${\sin}^2 x+a\cos x+{a}^2>1+\cos x$ holds for all real $x$

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 $$ \Huge 2 $$

We need that the inequality $\cos^2x+(1-a)\cos{x}-a^2<0$ will be true for all real $x$.

Let $\cos{x}=t$.

Thus, we need to find a smallest natural $a$ for which the inequality $$t^2+(1-a)t-a^2<0$$ is true for all $t\in[-1,1],$ for which we need $$(-1)^2+(1-a)(-1)-a^2<0$$ and $$1^2+(1-a)\cdot1-a^2<0,$$ which is $$a^2+a-2>0$$ and $$a^2-a>0,$$ which gives $$a\in(-\infty,-2)\cup(1,+\infty)$$ and we got the answer: $2$.

Write sin$^2$x=1-cos$^2$x, and factorize the resultant inequation.