We can write \begin{align*} f(x) &= 2 \cos^2 x - 1 - 2a (1 + \cos x) \\ &= 2 \cos^2 x - 2a \cos x - 1 - 2a \\ &= 2 \left( \cos x - \frac{a}{2} \right)^2 - \frac{1}{2} a^2 - 2a - 1.\end{align*}If $ a > 2, $ then $ f(x) $ attains its minimum value when $ \cos x = 1, $ in which case \[f(x) = 2 - 2a - 1 - 2a = 1 - 4a.\]If $ 1 - 4a = -\frac{1}{2}, $ then $ a = \frac{3}{8}, $ contradiction. If $ a < -2, $ then $ f(x) $ attains its minimum value when $ \cos x = -1, $ in which case \[f(x) = 2 + 2a - 1 - 2a = 1,\]so this case is not possible either. Otherwise, $ -2 \le a \le 2, $ and $ f(x) $ attains its minimum when $ \cos x = \frac{a}{2}, $ in which case \[f(x) = -\frac{1}{2} a^2 - 2a - 1.\]Thus, $ -\frac{1}{2} a^2 - 2a - 1 = -\frac{1}{2}, $ so $ a^2 + 4a + 1 = 0 $. By the quadratic formula, \[a = -2 \pm \sqrt{3}.\]Since $ -2 \le a \le 2, $ $ a = \boxed{-2 + \sqrt{3}} $.