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Let $sin^{2}(x) = 1 - cos^{2}(x)$ and substitute this identity into our original inequality:

$1 - cos^{2}(x) + a \cdot cos(x) + a^2 \ge 1 + cos(x) \Rightarrow 0 \ge cos^{2}(x) + (1-a)\cdot cos(x) - a^2$ (i)

Now let $u = cos(x)$ so that we have a concave-up parabola that satisfies (i) iff $u$ lies between its two roots. By the Quadratic Formula:

$u = \frac{(a-1) \pm \sqrt{(1-a)^2 - 4(1)(-a^2)}}{2} = \frac{(a-1) \pm \sqrt{5a^2 - 2a + 1}}{2} = cos(x)$ ;

or $\frac{(a-1) - \sqrt{5a^2 - 2a + 1}}{2} \le cos(x) \le \frac{(a-1) + \sqrt{5a^2 - 2a + 1}}{2}$ (ii).

Knowing that $-1 \le cos(x) \le 1$ for all $x \in \mathbb{R}$ , we can solve for $a$ in each bound of (ii):

$\frac{(a-1) - \sqrt{5a^2 - 2a + 1}}{2} = -1 \Rightarrow a+1 = \sqrt{5a^2 - 2a + 1} \Rightarrow a^2 + 2a + 1 = 5a^2 - 2a + 1 \Rightarrow 0 = 4a^2 - 4a \Rightarrow \boxed{a = 0, 1}.$ (iii)

$\frac{(a-1) + \sqrt{5a^2 - 2a + 1}}{2} = 1 \Rightarrow 3-a = \sqrt{5a^2 - 2a + 1} \Rightarrow 9 - 6a + a^2 = 5a^2 - 2a + 1 \Rightarrow 0 = 4a^2 + 4a - 8 = 4(a^2 + a - 2)$

or $0 = 4(a+2)(a-1) \Rightarrow \boxed{a = -2, 1}$ (iv)

We have three total integral values for $a$ of which (iii) and (iv) are both satisfied iff $a = 1$ . The number of integral values that $a$ cannot take is therefore two: $a = -2, 0.$