China Mathematical Competition

$$\begin{aligned} f(x) &= 2\cos^2 x - 1 - 2a - 2a \cos x \\ &= 2\left(\cos x - \frac{a}{2}\right)^2 - \frac{1}{2}a^2 - 2a - 1. \end{aligned}$$
For $a > 2$, $f(x)$ takes the minimum value of $1 - 4a$ when $\cos x = 1$; for $a < -2$, $f(x)$ takes the minimum $1$ when $\cos x = -1$; for $-2 \le a \le 2$, $f(x)$ takes the minimum $-\frac{1}{2}a^2 - 2a - 1$ when $\cos x = \frac{a}{2}$. It is easy to see that $f(x)$ will never be $-\frac{1}{2}$ for $a > 2$ or $a < -2$. So it is only possible that $-2 \le a \le 2$. Then from $-\frac{1}{2}a^2 - 2a - 1 = -\frac{1}{2}$, we get $a = -2 + \sqrt{3}$ or $a = -2 - \sqrt{3}$ (discarded). Therefore, the correct answer is $a = -2 + \sqrt{3}$.