9. (16 points) Let the real number $t \in [0, \pi]$. If the equation $\cos (x+t)=1-\cos x$ has a solution for $x$, find the range of values for $t$.


II. 9. The original equation is equivalent to
$$
\cos \left(x+\frac{t}{2}\right) \cdot \cos \frac{t}{2}=\frac{1}{2} \text {. }
$$

When $t=\pi$, the left side of equation (1) is 0, which clearly has no solution. When $t \in[0, \pi)$, equation (1) further simplifies to $\cos \left(x+\frac{t}{2}\right)=\frac{1}{2 \cos \frac{t}{2}}$.
Note that $\cos \left(x+\frac{t}{2}\right) \in[-1,1]$, and $\cos \frac{t}{2}>0$. Therefore, equation (1) has a solution if and only if $0<\frac{1}{2 \cos \frac{t}{2}} \leqslant 1$,
which implies $\frac{1}{2} \leqslant \cos \frac{t}{2} \leqslant 1 \Rightarrow 0 \leqslant t \leqslant \frac{\pi}{3}$.
In summary, $t \in\left[0, \frac{2 \pi}{3}\right]$.