China Mathematical Competition

$$\begin{array}{rl}f(x)& =2{\cos }^{2}x-1-2a-2a\cos x\\ & =2{\left(\cos x-\frac{a}{2}\right)}^2-\frac{1}{2}{a}^2-2a-1.\end{array}$$ 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 $a=-2+\sqrt{3}$.