If cos x = k has exactly one solution in [0, 2π],

Given: $\cos x=k$

If $k=0$, then

$\cos x=0$

$\Rightarrow \cos x=\cos \frac{\pi}{2}$

$\Rightarrow x=(2 n+1) \frac{\pi}{2}, n \in Z$

Now, $x=\frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}, \ldots$ for $n=1,2,3, \ldots$

If $k=1$, then

$\cos x=1$

$\Rightarrow \cos x=\cos 0$

$\Rightarrow x=2 m \pi, \mathrm{m} \in \mathrm{Z}$

Now, $x=2 \pi, 4 \pi, 6 \pi, 8 \pi, \ldots$ for $m=1,2,3,4, \ldots$

If $k=-1$, then

$\cos x=-1$

$\Rightarrow \cos x=\cos \pi$

$\Rightarrow x=2 \mathrm{p} \pi \pm \pi, \mathrm{p} \in \mathrm{Z}$

Now,

$x=2 p \pi+\pi$, i. e., $x=3 \pi, 5 \pi, 7 \pi, \ldots$ when $p=1,2,3, \ldots$

And,

$x=2 p \pi-\pi$, i. e., $x=\pi, 3 \pi, 5 \pi, 7 \pi, \ldots$ when $p=1,2,3,4, \ldots$

Clearly, we can see that for $x=\pi, \cos x=k$ has exactly one solution.

$\therefore k=-1$